I know it's possible to match a word and then reverse the matches using other tools (e.g. grep -v). However, is it possible to match lines that do not contain a specific word, e.g. hede, using a regular expression?
Input:
hoho
hihi
haha
hede
Code:
grep "<Regex for 'doesn't contain hede'>" input
Desired output:
hoho
hihi
haha
The notion that regex doesn't support inverse matching is not entirely true. You can mimic this behavior by using negative look-arounds:
^((?!hede).)*$
The regex above will match any string, or line without a line break, not containing the (sub)string 'hede'. As mentioned, this is not something regex is "good" at (or should do), but still, it is possible.
And if you need to match line break chars as well, use the DOT-ALL modifier (the trailing s in the following pattern):
/^((?!hede).)*$/s
or use it inline:
/(?s)^((?!hede).)*$/
(where the /.../ are the regex delimiters, i.e., not part of the pattern)
If the DOT-ALL modifier is not available, you can mimic the same behavior with the character class [\s\S]:
/^((?!hede)[\s\S])*$/
Explanation
A string is just a list of n characters. Before, and after each character, there's an empty string. So a list of n characters will have n+1 empty strings. Consider the string "ABhedeCD":
┌──┬───┬──┬───┬──┬───┬──┬───┬──┬───┬──┬───┬──┬───┬──┬───┬──┐
S = │e1│ A │e2│ B │e3│ h │e4│ e │e5│ d │e6│ e │e7│ C │e8│ D │e9│
└──┴───┴──┴───┴──┴───┴──┴───┴──┴───┴──┴───┴──┴───┴──┴───┴──┘
index 0 1 2 3 4 5 6 7
where the e's are the empty strings. The regex (?!hede). looks ahead to see if there's no substring "hede" to be seen, and if that is the case (so something else is seen), then the . (dot) will match any character except a line break. Look-arounds are also called zero-width-assertions because they don't consume any characters. They only assert/validate something.
So, in my example, every empty string is first validated to see if there's no "hede" up ahead, before a character is consumed by the . (dot). The regex (?!hede). will do that only once, so it is wrapped in a group, and repeated zero or more times: ((?!hede).)*. Finally, the start- and end-of-input are anchored to make sure the entire input is consumed: ^((?!hede).)*$
As you can see, the input "ABhedeCD" will fail because on e3, the regex (?!hede) fails (there is "hede" up ahead!).
Note that the solution to does not start with “hede”:
^(?!hede).*$
is generally much more efficient than the solution to does not contain “hede”:
^((?!hede).)*$
The former checks for “hede” only at the input string’s first position, rather than at every position.
If you're just using it for grep, you can use grep -v hede to get all lines which do not contain hede.
ETA Oh, rereading the question, grep -v is probably what you meant by "tools options".
Answer:
^((?!hede).)*$
Explanation:
^the beginning of the string,
( group and capture to \1 (0 or more times (matching the most amount possible)),
(?! look ahead to see if there is not,
hede your string,
) end of look-ahead,
. any character except \n,
)* end of \1 (Note: because you are using a quantifier on this capture, only the LAST repetition of the captured pattern will be stored in \1)
$ before an optional \n, and the end of the string
The given answers are perfectly fine, just an academic point:
Regular Expressions in the meaning of theoretical computer sciences ARE NOT ABLE do it like this. For them it had to look something like this:
^([^h].*$)|(h([^e].*$|$))|(he([^h].*$|$))|(heh([^e].*$|$))|(hehe.+$)
This only does a FULL match. Doing it for sub-matches would even be more awkward.
If you want the regex test to only fail if the entire string matches, the following will work:
^(?!hede$).*
e.g. -- If you want to allow all values except "foo" (i.e. "foofoo", "barfoo", and "foobar" will pass, but "foo" will fail), use: ^(?!foo$).*
Of course, if you're checking for exact equality, a better general solution in this case is to check for string equality, i.e.
myStr !== 'foo'
You could even put the negation outside the test if you need any regex features (here, case insensitivity and range matching):
!/^[a-f]oo$/i.test(myStr)
The regex solution at the top of this answer may be helpful, however, in situations where a positive regex test is required (perhaps by an API).
FWIW, since regular languages (aka rational languages) are closed under complementation, it's always possible to find a regular expression (aka rational expression) that negates another expression. But not many tools implement this.
Vcsn supports this operator (which it denotes {c}, postfix).
You first define the type of your expressions: labels are letter (lal_char) to pick from a to z for instance (defining the alphabet when working with complementation is, of course, very important), and the "value" computed for each word is just a Boolean: true the word is accepted, false, rejected.
In Python:
In [5]: import vcsn
c = vcsn.context('lal_char(a-z), b')
c
Out[5]: {a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z} → 𝔹
then you enter your expression:
In [6]: e = c.expression('(hede){c}'); e
Out[6]: (hede)^c
convert this expression to an automaton:
In [7]: a = e.automaton(); a
finally, convert this automaton back to a simple expression.
In [8]: print(a.expression())
\e+h(\e+e(\e+d))+([^h]+h([^e]+e([^d]+d([^e]+e[^]))))[^]*
where + is usually denoted |, \e denotes the empty word, and [^] is usually written . (any character). So, with a bit of rewriting ()|h(ed?)?|([^h]|h([^e]|e([^d]|d([^e]|e.)))).*.
You can see this example here, and try Vcsn online there.
Here's a good explanation of why it's not easy to negate an arbitrary regex. I have to agree with the other answers, though: if this is anything other than a hypothetical question, then a regex is not the right choice here.
With negative lookahead, regular expression can match something not contains specific pattern. This is answered and explained by Bart Kiers. Great explanation!
However, with Bart Kiers' answer, the lookahead part will test 1 to 4 characters ahead while matching any single character. We can avoid this and let the lookahead part check out the whole text, ensure there is no 'hede', and then the normal part (.*) can eat the whole text all at one time.
Here is the improved regex:
/^(?!.*?hede).*$/
Note the (*?) lazy quantifier in the negative lookahead part is optional, you can use (*) greedy quantifier instead, depending on your data: if 'hede' does present and in the beginning half of the text, the lazy quantifier can be faster; otherwise, the greedy quantifier be faster. However if 'hede' does not present, both would be equal slow.
Here is the demo code.
For more information about lookahead, please check out the great article: Mastering Lookahead and Lookbehind.
Also, please check out RegexGen.js, a JavaScript Regular Expression Generator that helps to construct complex regular expressions. With RegexGen.js, you can construct the regex in a more readable way:
var _ = regexGen;
var regex = _(
_.startOfLine(),
_.anything().notContains( // match anything that not contains:
_.anything().lazy(), 'hede' // zero or more chars that followed by 'hede',
// i.e., anything contains 'hede'
),
_.endOfLine()
);
Benchmarks
I decided to evaluate some of the presented Options and compare their performance, as well as use some new Features.
Benchmarking on .NET Regex Engine: http://regexhero.net/tester/
Benchmark Text:
The first 7 lines should not match, since they contain the searched Expression, while the lower 7 lines should match!
Regex Hero is a real-time online Silverlight Regular Expression Tester.
XRegex Hero is a real-time online Silverlight Regular Expression Tester.
Regex HeroRegex HeroRegex HeroRegex HeroRegex Hero is a real-time online Silverlight Regular Expression Tester.
Regex Her Regex Her Regex Her Regex Her Regex Her Regex Her Regex Hero is a real-time online Silverlight Regular Expression Tester.
Regex Her is a real-time online Silverlight Regular Expression Tester.Regex Hero
egex Hero egex Hero egex Hero egex Hero egex Hero egex Hero Regex Hero is a real-time online Silverlight Regular Expression Tester.
RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRegex Hero is a real-time online Silverlight Regular Expression Tester.
Regex Her
egex Hero
egex Hero is a real-time online Silverlight Regular Expression Tester.
Regex Her is a real-time online Silverlight Regular Expression Tester.
Regex Her Regex Her Regex Her Regex Her Regex Her Regex Her is a real-time online Silverlight Regular Expression Tester.
Nobody is a real-time online Silverlight Regular Expression Tester.
Regex Her o egex Hero Regex Hero Reg ex Hero is a real-time online Silverlight Regular Expression Tester.
Results:
Results are Iterations per second as the median of 3 runs - Bigger Number = Better
01: ^((?!Regex Hero).)*$ 3.914 // Accepted Answer
02: ^(?:(?!Regex Hero).)*$ 5.034 // With Non-Capturing group
03: ^(?!.*?Regex Hero).* 7.356 // Lookahead at the beginning, if not found match everything
04: ^(?>[^R]+|R(?!egex Hero))*$ 6.137 // Lookahead only on the right first letter
05: ^(?>(?:.*?Regex Hero)?)^.*$ 7.426 // Match the word and check if you're still at linestart
06: ^(?(?=.*?Regex Hero)(?#fail)|.*)$ 7.371 // Logic Branch: Find Regex Hero? match nothing, else anything
P1: ^(?(?=.*?Regex Hero)(*FAIL)|(*ACCEPT)) ????? // Logic Branch in Perl - Quick FAIL
P2: .*?Regex Hero(*COMMIT)(*FAIL)|(*ACCEPT) ????? // Direct COMMIT & FAIL in Perl
Since .NET doesn't support action Verbs (*FAIL, etc.) I couldn't test the solutions P1 and P2.
Summary:
The overall most readable and performance-wise fastest solution seems to be 03 with a simple negative lookahead. This is also the fastest solution for JavaScript, since JS does not support the more advanced Regex Features for the other solutions.
Not regex, but I've found it logical and useful to use serial greps with pipe to eliminate noise.
eg. search an apache config file without all the comments-
grep -v '\#' /opt/lampp/etc/httpd.conf # this gives all the non-comment lines
and
grep -v '\#' /opt/lampp/etc/httpd.conf | grep -i dir
The logic of serial grep's is (not a comment) and (matches dir)
Since no one else has given a direct answer to the question that was asked, I'll do it.
The answer is that with POSIX grep, it's impossible to literally satisfy this request:
grep "<Regex for 'doesn't contain hede'>" input
The reason is that with no flags, POSIX grep is only required to work with Basic Regular Expressions (BREs), which are simply not powerful enough for accomplishing that task, because of lack of alternation in subexpressions. The only kind of alternation it supports involves providing multiple regular expressions separated by newlines, and that doesn't cover all regular languages, e.g. there's no finite collection of BREs that matches the same regular language as the extended regular expression (ERE) ^(ab|cd)*$.
However, GNU grep implements extensions that allow it. In particular, \| is the alternation operator in GNU's implementation of BREs. If your regular expression engine supports alternation, parentheses and the Kleene star, and is able to anchor to the beginning and end of the string, that's all you need for this approach. Note however that negative sets [^ ... ] are very convenient in addition to those, because otherwise, you need to replace them with an expression of the form (a|b|c| ... ) that lists every character that is not in the set, which is extremely tedious and overly long, even more so if the whole character set is Unicode.
Thanks to formal language theory, we get to see how such an expression looks like. With GNU grep, the answer would be something like:
grep "^\([^h]\|h\(h\|eh\|edh\)*\([^eh]\|e[^dh]\|ed[^eh]\)\)*\(\|h\(h\|eh\|edh\)*\(\|e\|ed\)\)$" input
(found with Grail and some further optimizations made by hand).
You can also use a tool that implements EREs, like egrep, to get rid of the backslashes, or equivalently, pass the -E flag to POSIX grep (although I was under the impression that the question required avoiding any flags to grep whatsoever):
egrep "^([^h]|h(h|eh|edh)*([^eh]|e[^dh]|ed[^eh]))*(|h(h|eh|edh)*(|e|ed))$" input
Here's a script to test it (note it generates a file testinput.txt in the current directory). Several of the expressions presented in other answers fail this test.
#!/bin/bash
REGEX="^\([^h]\|h\(h\|eh\|edh\)*\([^eh]\|e[^dh]\|ed[^eh]\)\)*\(\|h\(h\|eh\|edh\)*\(\|e\|ed\)\)$"
# First four lines as in OP's testcase.
cat > testinput.txt <<EOF
hoho
hihi
haha
hede
h
he
ah
head
ahead
ahed
aheda
ahede
hhede
hehede
hedhede
hehehehehehedehehe
hedecidedthat
EOF
diff -s -u <(grep -v hede testinput.txt) <(grep "$REGEX" testinput.txt)
In my system it prints:
Files /dev/fd/63 and /dev/fd/62 are identical
as expected.
For those interested in the details, the technique employed is to convert the regular expression that matches the word into a finite automaton, then invert the automaton by changing every acceptance state to non-acceptance and vice versa, and then converting the resulting FA back to a regular expression.
As everyone has noted, if your regular expression engine supports negative lookahead, the regular expression is much simpler. For example, with GNU grep:
grep -P '^((?!hede).)*$' input
However, this approach has the disadvantage that it requires a backtracking regular expression engine. This makes it unsuitable in installations that are using secure regular expression engines like RE2, which is one reason to prefer the generated approach in some circumstances.
Using Kendall Hopkins' excellent FormalTheory library, written in PHP, which provides a functionality similar to Grail, and a simplifier written by myself, I've been able to write an online generator of negative regular expressions given an input phrase (only alphanumeric and space characters currently supported, and the length is limited): http://www.formauri.es/personal/pgimeno/misc/non-match-regex/
For hede it outputs:
^([^h]|h(h|e(h|dh))*([^eh]|e([^dh]|d[^eh])))*(h(h|e(h|dh))*(ed?)?)?$
which is equivalent to the above.
with this, you avoid to test a lookahead on each positions:
/^(?:[^h]+|h++(?!ede))*+$/
equivalent to (for .net):
^(?>(?:[^h]+|h+(?!ede))*)$
Old answer:
/^(?>[^h]+|h+(?!ede))*$/
Aforementioned (?:(?!hede).)* is great because it can be anchored.
^(?:(?!hede).)*$ # A line without hede
foo(?:(?!hede).)*bar # foo followed by bar, without hede between them
But the following would suffice in this case:
^(?!.*hede) # A line without hede
This simplification is ready to have "AND" clauses added:
^(?!.*hede)(?=.*foo)(?=.*bar) # A line with foo and bar, but without hede
^(?!.*hede)(?=.*foo).*bar # Same
An, in my opinon, more readable variant of the top answer:
^(?!.*hede)
Basically, "match at the beginning of the line if and only if it does not have 'hede' in it" - so the requirement translated almost directly into regex.
Of course, it's possible to have multiple failure requirements:
^(?!.*(hede|hodo|hada))
Details: The ^ anchor ensures the regex engine doesn't retry the match at every location in the string, which would match every string.
The ^ anchor in the beginning is meant to represent the beginning of the line. The grep tool matches each line one at a time, in contexts where you're working with a multiline string, you can use the "m" flag:
/^(?!.*hede)/m # JavaScript syntax
or
(?m)^(?!.*hede) # Inline flag
Here's how I'd do it:
^[^h]*(h(?!ede)[^h]*)*$
Accurate and more efficient than the other answers. It implements Friedl's "unrolling-the-loop" efficiency technique and requires much less backtracking.
Another option is that to add a positive look-ahead and check if hede is anywhere in the input line, then we would negate that, with an expression similar to:
^(?!(?=.*\bhede\b)).*$
with word boundaries.
The expression is explained on the top right panel of regex101.com, if you wish to explore/simplify/modify it, and in this link, you can watch how it would match against some sample inputs, if you like.
RegEx Circuit
jex.im visualizes regular expressions:
If you want to match a character to negate a word similar to negate character class:
For example, a string:
<?
$str="aaa bbb4 aaa bbb7";
?>
Do not use:
<?
preg_match('/aaa[^bbb]+?bbb7/s', $str, $matches);
?>
Use:
<?
preg_match('/aaa(?:(?!bbb).)+?bbb7/s', $str, $matches);
?>
Notice "(?!bbb)." is neither lookbehind nor lookahead, it's lookcurrent, for example:
"(?=abc)abcde", "(?!abc)abcde"
The OP did not specify or Tag the post to indicate the context (programming language, editor, tool) the Regex will be used within.
For me, I sometimes need to do this while editing a file using Textpad.
Textpad supports some Regex, but does not support lookahead or lookbehind, so it takes a few steps.
If I am looking to retain all lines that Do NOT contain the string hede, I would do it like this:
1. Search/replace the entire file to add a unique "Tag" to the beginning of each line containing any text.
Search string:^(.)
Replace string:<##-unique-##>\1
Replace-all
2. Delete all lines that contain the string hede (replacement string is empty):
Search string:<##-unique-##>.*hede.*\n
Replace string:<nothing>
Replace-all
3. At this point, all remaining lines Do NOT contain the string hede. Remove the unique "Tag" from all lines (replacement string is empty):
Search string:<##-unique-##>
Replace string:<nothing>
Replace-all
Now you have the original text with all lines containing the string hede removed.
If I am looking to Do Something Else to only lines that Do NOT contain the string hede, I would do it like this:
1. Search/replace the entire file to add a unique "Tag" to the beginning of each line containing any text.
Search string:^(.)
Replace string:<##-unique-##>\1
Replace-all
2. For all lines that contain the string hede, remove the unique "Tag":
Search string:<##-unique-##>(.*hede)
Replace string:\1
Replace-all
3. At this point, all lines that begin with the unique "Tag", Do NOT contain the string hede. I can now do my Something Else to only those lines.
4. When I am done, I remove the unique "Tag" from all lines (replacement string is empty):
Search string:<##-unique-##>
Replace string:<nothing>
Replace-all
Since the introduction of ruby-2.4.1, we can use the new Absent Operator in Ruby’s Regular Expressions
from the official doc
(?~abc) matches: "", "ab", "aab", "cccc", etc.
It doesn't match: "abc", "aabc", "ccccabc", etc.
Thus, in your case ^(?~hede)$ does the job for you
2.4.1 :016 > ["hoho", "hihi", "haha", "hede"].select{|s| /^(?~hede)$/.match(s)}
=> ["hoho", "hihi", "haha"]
Through PCRE verb (*SKIP)(*F)
^hede$(*SKIP)(*F)|^.*$
This would completely skips the line which contains the exact string hede and matches all the remaining lines.
DEMO
Execution of the parts:
Let us consider the above regex by splitting it into two parts.
Part before the | symbol. Part shouldn't be matched.
^hede$(*SKIP)(*F)
Part after the | symbol. Part should be matched.
^.*$
PART 1
Regex engine will start its execution from the first part.
^hede$(*SKIP)(*F)
Explanation:
^ Asserts that we are at the start.
hede Matches the string hede
$ Asserts that we are at the line end.
So the line which contains the string hede would be matched. Once the regex engine sees the following (*SKIP)(*F) (Note: You could write (*F) as (*FAIL)) verb, it skips and make the match to fail. | called alteration or logical OR operator added next to the PCRE verb which inturn matches all the boundaries exists between each and every character on all the lines except the line contains the exact string hede. See the demo here. That is, it tries to match the characters from the remaining string. Now the regex in the second part would be executed.
PART 2
^.*$
Explanation:
^ Asserts that we are at the start. ie, it matches all the line starts except the one in the hede line. See the demo here.
.* In the Multiline mode, . would match any character except newline or carriage return characters. And * would repeat the previous character zero or more times. So .* would match the whole line. See the demo here.
Hey why you added .* instead of .+ ?
Because .* would match a blank line but .+ won't match a blank. We want to match all the lines except hede , there may be a possibility of blank lines also in the input . so you must use .* instead of .+ . .+ would repeat the previous character one or more times. See .* matches a blank line here.
$ End of the line anchor is not necessary here.
The TXR Language supports regex negation.
$ txr -c '#(repeat)
#{nothede /~hede/}
#(do (put-line nothede))
#(end)' Input
A more complicated example: match all lines that start with a and end with z, but do not contain the substring hede:
$ txr -c '#(repeat)
#{nothede /a.*z&~.*hede.*/}
#(do (put-line nothede))
#(end)' -
az <- echoed
az
abcz <- echoed
abcz
abhederz <- not echoed; contains hede
ahedez <- not echoed; contains hede
ace <- not echoed; does not end in z
ahedz <- echoed
ahedz
Regex negation is not particularly useful on its own but when you also have intersection, things get interesting, since you have a full set of boolean set operations: you can express "the set which matches this, except for things which match that".
It may be more maintainable to two regexes in your code, one to do the first match, and then if it matches run the second regex to check for outlier cases you wish to block for example ^.*(hede).* then have appropriate logic in your code.
OK, I admit this is not really an answer to the posted question posted and it may also use slightly more processing than a single regex. But for developers who came here looking for a fast emergency fix for an outlier case then this solution should not be overlooked.
The below function will help you get your desired output
<?PHP
function removePrepositions($text){
$propositions=array('/\bfor\b/i','/\bthe\b/i');
if( count($propositions) > 0 ) {
foreach($propositions as $exceptionPhrase) {
$text = preg_replace($exceptionPhrase, '', trim($text));
}
$retval = trim($text);
}
return $retval;
}
?>
I wanted to add another example for if you are trying to match an entire line that contains string X, but does not also contain string Y.
For example, let's say we want to check if our URL / string contains "tasty-treats", so long as it does not also contain "chocolate" anywhere.
This regex pattern would work (works in JavaScript too)
^(?=.*?tasty-treats)((?!chocolate).)*$
(global, multiline flags in example)
Interactive Example: https://regexr.com/53gv4
Matches
(These urls contain "tasty-treats" and also do not contain "chocolate")
example.com/tasty-treats/strawberry-ice-cream
example.com/desserts/tasty-treats/banana-pudding
example.com/tasty-treats-overview
Does Not Match
(These urls contain "chocolate" somewhere - so they won't match even though they contain "tasty-treats")
example.com/tasty-treats/chocolate-cake
example.com/home-cooking/oven-roasted-chicken
example.com/tasty-treats/banana-chocolate-fudge
example.com/desserts/chocolate/tasty-treats
example.com/chocolate/tasty-treats/desserts
As long as you are dealing with lines, simply mark the negative matches and target the rest.
In fact, I use this trick with sed because ^((?!hede).)*$ looks not supported by it.
For the desired output
Mark the negative match: (e.g. lines with hede), using a character not included in the whole text at all. An emoji could probably be a good choice for this purpose.
s/(.*hede)/🔒\1/g
Target the rest (the unmarked strings: e.g. lines without hede). Suppose you want to keep only the target and delete the rest (as you want):
s/^🔒.*//g
For a better understanding
Suppose you want to delete the target:
Mark the negative match: (e.g. lines with hede), using a character not included in the whole text at all. An emoji could probably be a good choice for this purpose.
s/(.*hede)/🔒\1/g
Target the rest (the unmarked strings: e.g. lines without hede). Suppose you want to delete the target:
s/^[^🔒].*//g
Remove the mark:
s/🔒//g
^((?!hede).)*$ is an elegant solution, except since it consumes characters you won't be able to combine it with other criteria. For instance, say you wanted to check for the non-presence of "hede" and the presence of "haha." This solution would work because it won't consume characters:
^(?!.*\bhede\b)(?=.*\bhaha\b)
How to use PCRE's backtracking control verbs to match a line not containing a word
Here's a method that I haven't seen used before:
/.*hede(*COMMIT)^|/
How it works
First, it tries to find "hede" somewhere in the line. If successful, at this point, (*COMMIT) tells the engine to, not only not backtrack in the event of a failure, but also not to attempt any further matching in that case. Then, we try to match something that cannot possibly match (in this case, ^).
If a line does not contain "hede" then the second alternative, an empty subpattern, successfully matches the subject string.
This method is no more efficient than a negative lookahead, but I figured I'd just throw it on here in case someone finds it nifty and finds a use for it for other, more interesting applications.
Simplest thing that I could find would be
[^(hede)]
Tested at https://regex101.com/
You can also add unit-test cases on that site
A simpler solution is to use the not operator !
Your if statement will need to match "contains" and not match "excludes".
var contains = /abc/;
var excludes =/hede/;
if(string.match(contains) && !(string.match(excludes))){ //proceed...
I believe the designers of RegEx anticipated the use of not operators.
Related
I have a value like this:
"Foo Bar" "Another Value" something else
What regex will return the values enclosed in the quotation marks (e.g. Foo Bar and Another Value)?
In general, the following regular expression fragment is what you are looking for:
"(.*?)"
This uses the non-greedy *? operator to capture everything up to but not including the next double quote. Then, you use a language-specific mechanism to extract the matched text.
In Python, you could do:
>>> import re
>>> string = '"Foo Bar" "Another Value"'
>>> print re.findall(r'"(.*?)"', string)
['Foo Bar', 'Another Value']
I've been using the following with great success:
(["'])(?:(?=(\\?))\2.)*?\1
It supports nested quotes as well.
For those who want a deeper explanation of how this works, here's an explanation from user ephemient:
([""']) match a quote; ((?=(\\?))\2.) if backslash exists, gobble it, and whether or not that happens, match a character; *? match many times (non-greedily, as to not eat the closing quote); \1 match the same quote that was use for opening.
I would go for:
"([^"]*)"
The [^"] is regex for any character except '"'
The reason I use this over the non greedy many operator is that I have to keep looking that up just to make sure I get it correct.
Lets see two efficient ways that deal with escaped quotes. These patterns are not designed to be concise nor aesthetic, but to be efficient.
These ways use the first character discrimination to quickly find quotes in the string without the cost of an alternation. (The idea is to discard quickly characters that are not quotes without to test the two branches of the alternation.)
Content between quotes is described with an unrolled loop (instead of a repeated alternation) to be more efficient too: [^"\\]*(?:\\.[^"\\]*)*
Obviously to deal with strings that haven't balanced quotes, you can use possessive quantifiers instead: [^"\\]*+(?:\\.[^"\\]*)*+ or a workaround to emulate them, to prevent too much backtracking. You can choose too that a quoted part can be an opening quote until the next (non-escaped) quote or the end of the string. In this case there is no need to use possessive quantifiers, you only need to make the last quote optional.
Notice: sometimes quotes are not escaped with a backslash but by repeating the quote. In this case the content subpattern looks like this: [^"]*(?:""[^"]*)*
The patterns avoid the use of a capture group and a backreference (I mean something like (["']).....\1) and use a simple alternation but with ["'] at the beginning, in factor.
Perl like:
["'](?:(?<=")[^"\\]*(?s:\\.[^"\\]*)*"|(?<=')[^'\\]*(?s:\\.[^'\\]*)*')
(note that (?s:...) is a syntactic sugar to switch on the dotall/singleline mode inside the non-capturing group. If this syntax is not supported you can easily switch this mode on for all the pattern or replace the dot with [\s\S])
(The way this pattern is written is totally "hand-driven" and doesn't take account of eventual engine internal optimizations)
ECMA script:
(?=["'])(?:"[^"\\]*(?:\\[\s\S][^"\\]*)*"|'[^'\\]*(?:\\[\s\S][^'\\]*)*')
POSIX extended:
"[^"\\]*(\\(.|\n)[^"\\]*)*"|'[^'\\]*(\\(.|\n)[^'\\]*)*'
or simply:
"([^"\\]|\\.|\\\n)*"|'([^'\\]|\\.|\\\n)*'
Peculiarly, none of these answers produce a regex where the returned match is the text inside the quotes, which is what is asked for. MA-Madden tries but only gets the inside match as a captured group rather than the whole match. One way to actually do it would be :
(?<=(["']\b))(?:(?=(\\?))\2.)*?(?=\1)
Examples for this can be seen in this demo https://regex101.com/r/Hbj8aP/1
The key here is the the positive lookbehind at the start (the ?<= ) and the positive lookahead at the end (the ?=). The lookbehind is looking behind the current character to check for a quote, if found then start from there and then the lookahead is checking the character ahead for a quote and if found stop on that character. The lookbehind group (the ["']) is wrapped in brackets to create a group for whichever quote was found at the start, this is then used at the end lookahead (?=\1) to make sure it only stops when it finds the corresponding quote.
The only other complication is that because the lookahead doesn't actually consume the end quote, it will be found again by the starting lookbehind which causes text between ending and starting quotes on the same line to be matched. Putting a word boundary on the opening quote (["']\b) helps with this, though ideally I'd like to move past the lookahead but I don't think that is possible. The bit allowing escaped characters in the middle I've taken directly from Adam's answer.
The RegEx of accepted answer returns the values including their sourrounding quotation marks: "Foo Bar" and "Another Value" as matches.
Here are RegEx which return only the values between quotation marks (as the questioner was asking for):
Double quotes only (use value of capture group #1):
"(.*?[^\\])"
Single quotes only (use value of capture group #1):
'(.*?[^\\])'
Both (use value of capture group #2):
(["'])(.*?[^\\])\1
-
All support escaped and nested quotes.
I liked Eugen Mihailescu's solution to match the content between quotes whilst allowing to escape quotes. However, I discovered some problems with escaping and came up with the following regex to fix them:
(['"])(?:(?!\1|\\).|\\.)*\1
It does the trick and is still pretty simple and easy to maintain.
Demo (with some more test-cases; feel free to use it and expand on it).
PS: If you just want the content between quotes in the full match ($0), and are not afraid of the performance penalty use:
(?<=(['"])\b)(?:(?!\1|\\).|\\.)*(?=\1)
Unfortunately, without the quotes as anchors, I had to add a boundary \b which does not play well with spaces and non-word boundary characters after the starting quote.
Alternatively, modify the initial version by simply adding a group and extract the string form $2:
(['"])((?:(?!\1|\\).|\\.)*)\1
PPS: If your focus is solely on efficiency, go with Casimir et Hippolyte's solution; it's a good one.
A very late answer, but like to answer
(\"[\w\s]+\")
http://regex101.com/r/cB0kB8/1
The pattern (["'])(?:(?=(\\?))\2.)*?\1 above does the job but I am concerned of its performances (it's not bad but could be better). Mine below it's ~20% faster.
The pattern "(.*?)" is just incomplete. My advice for everyone reading this is just DON'T USE IT!!!
For instance it cannot capture many strings (if needed I can provide an exhaustive test-case) like the one below:
$string = 'How are you? I\'m fine, thank you';
The rest of them are just as "good" as the one above.
If you really care both about performance and precision then start with the one below:
/(['"])((\\\1|.)*?)\1/gm
In my tests it covered every string I met but if you find something that doesn't work I would gladly update it for you.
Check my pattern in an online regex tester.
This version
accounts for escaped quotes
controls backtracking
/(["'])((?:(?!\1)[^\\]|(?:\\\\)*\\[^\\])*)\1/
MORE ANSWERS! Here is the solution i used
\"([^\"]*?icon[^\"]*?)\"
TLDR;
replace the word icon with what your looking for in said quotes and voila!
The way this works is it looks for the keyword and doesn't care what else in between the quotes.
EG:
id="fb-icon"
id="icon-close"
id="large-icon-close"
the regex looks for a quote mark "
then it looks for any possible group of letters thats not "
until it finds icon
and any possible group of letters that is not "
it then looks for a closing "
I liked Axeman's more expansive version, but had some trouble with it (it didn't match for example
foo "string \\ string" bar
or
foo "string1" bar "string2"
correctly, so I tried to fix it:
# opening quote
(["'])
(
# repeat (non-greedy, so we don't span multiple strings)
(?:
# anything, except not the opening quote, and not
# a backslash, which are handled separately.
(?!\1)[^\\]
|
# consume any double backslash (unnecessary?)
(?:\\\\)*
|
# Allow backslash to escape characters
\\.
)*?
)
# same character as opening quote
\1
string = "\" foo bar\" \"loloo\""
print re.findall(r'"(.*?)"',string)
just try this out , works like a charm !!!
\ indicates skip character
Unlike Adam's answer, I have a simple but worked one:
(["'])(?:\\\1|.)*?\1
And just add parenthesis if you want to get content in quotes like this:
(["'])((?:\\\1|.)*?)\1
Then $1 matches quote char and $2 matches content string.
All the answer above are good.... except they DOES NOT support all the unicode characters! at ECMA Script (Javascript)
If you are a Node users, you might want the the modified version of accepted answer that support all unicode characters :
/(?<=((?<=[\s,.:;"']|^)["']))(?:(?=(\\?))\2.)*?(?=\1)/gmu
Try here.
My solution to this is below
(["']).*\1(?![^\s])
Demo link : https://regex101.com/r/jlhQhV/1
Explanation:
(["'])-> Matches to either ' or " and store it in the backreference \1 once the match found
.* -> Greedy approach to continue matching everything zero or more times until it encounters ' or " at end of the string. After encountering such state, regex engine backtrack to previous matching character and here regex is over and will move to next regex.
\1 -> Matches to the character or string that have been matched earlier with the first capture group.
(?![^\s]) -> Negative lookahead to ensure there should not any non space character after the previous match
echo 'junk "Foo Bar" not empty one "" this "but this" and this neither' | sed 's/[^\"]*\"\([^\"]*\)\"[^\"]*/>\1</g'
This will result in: >Foo Bar<><>but this<
Here I showed the result string between ><'s for clarity, also using the non-greedy version with this sed command we first throw out the junk before and after that ""'s and then replace this with the part between the ""'s and surround this by ><'s.
From Greg H. I was able to create this regex to suit my needs.
I needed to match a specific value that was qualified by being inside quotes. It must be a full match, no partial matching could should trigger a hit
e.g. "test" could not match for "test2".
reg = r"""(['"])(%s)\1"""
if re.search(reg%(needle), haystack, re.IGNORECASE):
print "winning..."
Hunter
If you're trying to find strings that only have a certain suffix, such as dot syntax, you can try this:
\"([^\"]*?[^\"]*?)\".localized
Where .localized is the suffix.
Example:
print("this is something I need to return".localized + "so is this".localized + "but this is not")
It will capture "this is something I need to return".localized and "so is this".localized but not "but this is not".
A supplementary answer for the subset of Microsoft VBA coders only one uses the library Microsoft VBScript Regular Expressions 5.5 and this gives the following code
Sub TestRegularExpression()
Dim oRE As VBScript_RegExp_55.RegExp '* Tools->References: Microsoft VBScript Regular Expressions 5.5
Set oRE = New VBScript_RegExp_55.RegExp
oRE.Pattern = """([^""]*)"""
oRE.Global = True
Dim sTest As String
sTest = """Foo Bar"" ""Another Value"" something else"
Debug.Assert oRE.test(sTest)
Dim oMatchCol As VBScript_RegExp_55.MatchCollection
Set oMatchCol = oRE.Execute(sTest)
Debug.Assert oMatchCol.Count = 2
Dim oMatch As Match
For Each oMatch In oMatchCol
Debug.Print oMatch.SubMatches(0)
Next oMatch
End Sub
I had to match a number followed by itself 14 times. Then I've came to the following regular expression in the regexstor.net/tester:
(\d)\1{14}
Edit
When I paste it in my code, including the backslashes properly:
"(\\d)\\1{14}"
I've replaced the back-reference "\1" by the "$1" which is used to replace matches in Java.
Then I've realized that it doesn't work. When you need to back-reference a match in the REGEX, in Java, you have to use "\N", but when you want to replace it, the operator is "$N".
My question is: why?
$1 is not a back reference in Java's regexes, nor in any other flavor I can think of. You only use $1 when you are replacing something:
String input="A12.3 bla bla my input";
input = StringUtils.replacePattern(
input, "^([A-Z]\\d{2}\\.\\d).*$", "$1");
// ^^^^
There is some misinformation about what a back reference is, including the very place I got that snippet from: simple java regex with backreference does not work.
Java modeled its regex syntax after other existing flavors where the $ was already a meta character. It anchors to the end of the string (or line in multi-line mode).
Similarly, Java uses \1 for back references. Because regexes are strings, it must be escaped: \\1.
From a lexical/syntactic standpoint it is true that $1 could be used unambiguously (as a bonus it would prevent the need for the "evil escaped escape" when using back references).
To match a 1 that comes after the end of a line the regex would need to be $\n1:
this line
1
It just makes more sense to use a familiar syntax instead of changing the rules, most of which came from Perl.
The first version of Perl came out in 1987, which is much earlier than Java, which was released in beta in 1995.
I dug up the man pages for Perl 1, which say:
The bracketing construct (\ ...\ ) may also be used, in which case \<digit> matches the digit'th substring. (Outside of the pattern, always use $ instead of \ in front of the digit. The scope of $<digit> (and $\`, $& and $') extends to the end of the enclosing BLOCK or eval string, or to the next pattern match with subexpressions. The \<digit> notation sometimes works outside the current pattern, but should not be relied upon.) You may have as many parentheses as you wish. If you have more than 9 substrings, the variables $10, $11, ... refer to the corresponding substring. Within the pattern, \10, \11, etc. refer back to substrings if there have been at least that many left parens before the backreference. Otherwise (for backward compatibilty) \10 is the same as \010, a backspace, and \11 the same as \011, a tab. And so on. (\1 through \9 are always backreferences.)
I think the main Problem is not the backreference - which works perfectly fine with \1 in java.
Your Problem is more likely the "overall" escaping of a regex pattern in Java.
If you want to have the pattern
(\d)\1{14}
passed to the regex engine, you first need to escape it cause it's a java-string when you write it:
(\\d)\\1{14}
Voila, works like a charm: goo.gl/BNCx7B (add http://, SO does not allow Url-Shorteners, but tutorialspoint.com has no other option as it seems)
Offline-Example:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class HelloWorld{
public static void main(String []args){
String test = "555555555555555"; // 5 followed by 5 for 14 times.
String pattern = "(\\d)\\1{14}";
Pattern r = Pattern.compile(pattern);
Matcher m = r.matcher(test);
if (m.find( )) {
System.out.println("Matched!");
}else{
System.out.println("not matched :-(");
}
}
}
I had to match a number followed by itself 14 times. Then I've came to the following regular expression in the regexstor.net/tester:
(\d)\1{14}
Edit
When I paste it in my code, including the backslashes properly:
"(\\d)\\1{14}"
I've replaced the back-reference "\1" by the "$1" which is used to replace matches in Java.
Then I've realized that it doesn't work. When you need to back-reference a match in the REGEX, in Java, you have to use "\N", but when you want to replace it, the operator is "$N".
My question is: why?
$1 is not a back reference in Java's regexes, nor in any other flavor I can think of. You only use $1 when you are replacing something:
String input="A12.3 bla bla my input";
input = StringUtils.replacePattern(
input, "^([A-Z]\\d{2}\\.\\d).*$", "$1");
// ^^^^
There is some misinformation about what a back reference is, including the very place I got that snippet from: simple java regex with backreference does not work.
Java modeled its regex syntax after other existing flavors where the $ was already a meta character. It anchors to the end of the string (or line in multi-line mode).
Similarly, Java uses \1 for back references. Because regexes are strings, it must be escaped: \\1.
From a lexical/syntactic standpoint it is true that $1 could be used unambiguously (as a bonus it would prevent the need for the "evil escaped escape" when using back references).
To match a 1 that comes after the end of a line the regex would need to be $\n1:
this line
1
It just makes more sense to use a familiar syntax instead of changing the rules, most of which came from Perl.
The first version of Perl came out in 1987, which is much earlier than Java, which was released in beta in 1995.
I dug up the man pages for Perl 1, which say:
The bracketing construct (\ ...\ ) may also be used, in which case \<digit> matches the digit'th substring. (Outside of the pattern, always use $ instead of \ in front of the digit. The scope of $<digit> (and $\`, $& and $') extends to the end of the enclosing BLOCK or eval string, or to the next pattern match with subexpressions. The \<digit> notation sometimes works outside the current pattern, but should not be relied upon.) You may have as many parentheses as you wish. If you have more than 9 substrings, the variables $10, $11, ... refer to the corresponding substring. Within the pattern, \10, \11, etc. refer back to substrings if there have been at least that many left parens before the backreference. Otherwise (for backward compatibilty) \10 is the same as \010, a backspace, and \11 the same as \011, a tab. And so on. (\1 through \9 are always backreferences.)
I think the main Problem is not the backreference - which works perfectly fine with \1 in java.
Your Problem is more likely the "overall" escaping of a regex pattern in Java.
If you want to have the pattern
(\d)\1{14}
passed to the regex engine, you first need to escape it cause it's a java-string when you write it:
(\\d)\\1{14}
Voila, works like a charm: goo.gl/BNCx7B (add http://, SO does not allow Url-Shorteners, but tutorialspoint.com has no other option as it seems)
Offline-Example:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class HelloWorld{
public static void main(String []args){
String test = "555555555555555"; // 5 followed by 5 for 14 times.
String pattern = "(\\d)\\1{14}";
Pattern r = Pattern.compile(pattern);
Matcher m = r.matcher(test);
if (m.find( )) {
System.out.println("Matched!");
}else{
System.out.println("not matched :-(");
}
}
}
I am trying to figure out a regex to match a password that contains
one upper case letter.
one number
one special character.
and at least 4 characters of length
the regex that I wrote is
^((?=.*[0-9])(?=.*[A-Z])(?=.*[^A-Za-z0-9])){4,}
however it is not working, and I couldn't figure out why.
So please can someone tell me why this code is not working, where did I mess up, and how to correct this code.
Your regex can be rewritten as
^(
(?=.*[0-9])
(?=.*[A-Z])
(?=.*[^A-Za-z0-9])
){4,}
As you see {4,} applies to group which doesn't let you match any character since look-around is zero-width, which effectively means "4 or more of nothing".
You need to add . before {4,} to let your regex handle "and at least 4 characters of length" point (rest is handled by look-around).
You can remove that capturing group since you don't really need it.
So try with something like
^(?=.*[0-9])(?=.*[A-Z])(?=.*[^A-Za-z0-9]).{4,}
You could come up with sth. like:
^(?=.*[A-Z])(?=.*\d)(?=.*[!"§$%&/()=?`]).{4,}$
In multiline mode, see a demo on regex101.com.
This approach specifies the special characters directly (which could be extended, obviously).
From the following list only the bold ones would satisfy these criteria:
test
Test123!
StrongPassword34?
weakone
Tabaluga"12???
You can still enhance this expression by being more specific and requiring contrary pairs. Just to remind you, the dot-star (.*) brings you down the line and then backtracks eventually. This will almost always require more steps than to directly look for contrary pairs.
Consider the following expression:
^ # bind the expression to the beginning of the string
(?=[^A-Z\n\r]*[A-Z]) # look ahead for sth. that is not A-Z, or newline and require one of A-Z
(?=[^\d\n\r]*\d) # same construct for digits
(?=\w*[^\w\n\r]) # same construct for special chars (\w = _A-Za-z0-9)
.{4,}
$
You'll see a significant reduction in steps as the regex engine does not have to backtrack everytime.
I need to create a regular expression that allows a string to contain any number of:
alphanumeric characters
spaces
(
)
&
.
No other characters are permitted. I used RegexBuddy to construct the following regex, which works correctly when I test it within RegexBuddy:
\w* *\(*\)*&*\.*
Then I used RegexBuddy's "Use" feature to convert this into Java code, but it doesn't appear to work correctly using a simple test program:
public class RegexTest
{
public static void main(String[] args)
{
String test = "(AT) & (T)."; // Should be valid
System.out.println("Test string matches: "
+ test.matches("\\w* *\\(*\\)*&*\\.*")); // Outputs false
}
}
I must admit that I have a bit of a blind spot when it comes to regular expressions. Can anyone explain why it doesn't work please?
That regular expression tests for any amount of whitespace, followed by any amount of alphanumeric characters, followed by any amount of open parens, followed by any amount of close parens, followed by any amount of ampersands, followed by any amount of periods.
What you want is...
test.matches("[\\w \\(\\)&\\.]*")
As mentioned by mmyers, this allows the empty string. If you do not want to allow the empty string...
test.matches("[\\w \\(\\)&\\.]+")
Though that will also allow a string that is only spaces, or only periods, etc.. If you want to ensure at least one alpha-numeric character...
test.matches("[\\w \\(\\)&\\.]*\\w+[\\w \\(\\)&\\.]*")
So you understand what the regular expression is saying... anything within the square brackets ("[]") indicates a set of characters. So, where "a*" means 0 or more a's, [abc]* means 0 or more characters, all of which being a's, b's, or c's.
Maybe I'm misunderstanding your description, but aren't you essentially defining a class of characters without an order rather than a specific sequence? Shouldn't your regexp have a structure of [xxxx]+, where xxxx are the actual characters you want ?
The difference between your Java code snippet and the Test tab in RegexBuddy is that the matches() method in Java requires the regular expression to match the whole string, while the Test tab in RegexBuddy allows partial matches. If you use your original regex in RegexBuddy, you'll see multiple blocks of yellow and blue highlighting. That indicates RegexBuddy found multiple partial matches in your string. To get a regex that works as intended with matches(), you need to edit it until the whole test subject is highlighted in yellow, or if you turn off highlighting, until the Find First button selects the whole text.
Alternatively, you can use the anchors \A and \Z at the start and the end of your regex to force it to match the whole string. When you do that, your regex always behaves in the same way, whether you test it in RegexBuddy, or whether you use matches() or another method in Java. Only matches() requires a full string match. All other Matcher methods in Java allow partial matches.
the regex
\w* *\(*\)*&*\.*
will give you the items you described, but only in the order you described, and each one can be as many as wanted. So "skjhsklasdkjgsh((((())))))&&&&&....." works, but not mixing the characters.
You want a regex like this:
\[\w\(\)\&\.]+\
which will allow a mix of all characters.
edit: my regex knowledge is limited, so the above syntax may not be perfect.