Develop Reference

Modifying existing Java regex

I have the following regex that validates the allowed characters:
^[a-zA-Z0-9-?\/:;(){}\[\]|`~´.\,'+÷ !##$£%^"&*_<>=àáâäçèéêëìíîïñòóôöùúûüýßÀÁÂÄÇÈÉÊËÌÍÎÏÒÓÔÖÙÚÛÜÑ\\]*$
I need to modify it so that the string being validated:
may not begin with space or “/”
may not contain “//”
may not end with “/”
For the space at the beginning I have adapted it to
^[^\s][a-zA-Z0-9-?\\/:;(){}\\[\\]|`~´.\\,'+÷ !##$£%^\"&*_<>=àáâäçèéêëìíîïñòóôöùúûüýßÀÁÂÄÇÈÉÊËÌÍÎÏÒÓÔÖÙÚÛÜÑ\\\\]*$
Not sure what to do about the other two requirements
For the second one I tried combining it with ^((?!//))*$ in various ways but to no success.

Note that ^((?!\/\/))*$ matches any empty string since the lookahead is a non-consuming pattern and here it always returns true.
[^\s] at the start of your pattern will match any chars other than whitespace chars, even those you did not specify in the character class.
You can use
^(?![\s/])(?!.*//)[a-zA-Z0-9?/:;(){}\[\]|`~´.,'+÷ !##$£%^\"&*_<>=àáâäçèéêëìíîïñòóôöùúûüýßÀÁÂÄÇÈÉÊËÌÍÎÏÒÓÔÖÙÚÛÜÑ\\-]*$(?<!/)
See the regex demo. Details:
^(?![\s/])(?!.*//) - at the start of string, two checks are peformed:
(?![\s/]) - no whitespace or / allowed (right at the start)
(?!.*//) - no // allowed anywhere after zero or more chars other than line break chars, as many as possible
(?<!/) is the check after the end of string is hit, and it fails the match if the last char in string is /.
Note that in Java regex declarations, you do not need to escape / since regex delimiter notation is not used, and / itself is not a special regex metacharacter.

It seems like the following regexp should be enough and more simple: (?!.*//)^[^ /].*[^/]$
So at the beginning you can use negative lookahead to prevent occurence of // anywhere in the text. Then any character but space and / is accepted at the beginning, then anything can be present (besides // which was excluded by negative lookahead) and anything but / is accepted at the end.

Since 95% of the time the special conditions on the space and forward slash
will not occur, it might be better to take those two characters out of your
big class and handle them separately if and when they occur.
The big class can also be condensed to speed things up a bit.
^(?>[a-zA-Z0-9\\!-.:-#\[\]-`{-~£´ÄÖäö÷À-ÂÇ-ÏÑ-ÔÙ-Üß-âç-ïñ-ôù-ý]+|(?:/(?!/|$)|[ ])(?<!^.))*$
https://regex101.com/r/LpCwt6/1
^
(?>
[a-zA-Z0-9\\!-.:-#\[\]-`{-~£´ÄÖäö÷À-ÂÇ-ÏÑ-ÔÙ-Üß-âç-ïñ-ôù-ý]+
| (?:
/
(?! / | $ )
| [ ]
)
(?<! ^ . )
)*
$
And if you want to absorb all the class characters it can get very small.
^(?>[!-.0-~£´ÄÖäö÷À-ÂÇ-ÏÑ-ÔÙ-Üß-âç-ïñ-ôù-ý]+|(?:/(?!/|$)|[ ])(?<!^.))*$
https://regex101.com/r/EYdM5C/1

Regular Expression to exclude a particular filename in Java [duplicate]

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.

Free space regex option (Pattern.COMMENTS) not working as expected

I'm trying to detect profanity using regex. But I want to detect the word even if they've spaced out the word like "Profa nity". However when using the "(?x)" option it still doesn't want to detect.
I currently got:
(?ix).*Bad Word.*
I've tried using http://www.rubular.com to debug the expression with not luck.
If it helps in any way it's for at Teamspeak Bot where I want to kick the user for having banned words in their name. In the config it refers to http://docs.oracle.com/javase/1.5.0/docs/api/java/util/regex/Pattern.html where I can't find anything relating to the (?) options.
The bot itself can be found here: https://forum.teamspeak.com/threads/51286-JTS3ServerMod-Multifunction-TS3-Server-Bot-(Idle-Record-Away-Mute-Welcome-)

when using the "(?x)" option it still doesn't want to detect
The (?x) is an embedded flag option (also known as an inline modifier/option) enables the Pattern.COMMENTS option, also known as free-spacing mode that enables comments inside regular expressions and makes the regex engine ignore all regular whitespace inside the pattern. As per Free-Spacing in Character Classes:
In free-spacing mode, whitespace between regular expression tokens is ignored. Whitespace includes spaces, tabs, and line breaks. Note that only whitespace between tokens is ignored. a b c is the same as abc in free-spacing mode. But \ d and \d are not the same. The former matches d, while the latter matches a digit. \d is a single regex token composed of a backslash and a "d". Breaking up the token with a space gives you an escaped space (which matches a space), and a literal "d".
Likewise, grouping modifiers cannot be broken up. (?>atomic) is the same as (?> ato mic ) and as ( ?>ato mic). They all match the same atomic group. They're not the same as (? >atomic). The latter is a syntax error. The ?> grouping modifier is a single element in the regex syntax, and must stay together. This is true for all such constructs, including lookaround, named groups, etc.
So, to match a single space in a pattern with the (?x) modifier, you need to escape it:
String reg = "(?ix).*Bad\\ Word.*"; // Escaped space matches a space in free spacing mode
String reg = "(?ix).* Bad\\ Word .*"; // More formatting spaces, same pattern
NOTE that you CAN'T put the space into a character class to make it meaningful in a Java regex. See below:
Java, however, does not treat a character class as a single token in free-spacing mode. Java does ignore spaces, line breaks, and comments inside character classes. So in Java's free-spacing mode, [abc] is identical to [ a b c ].
Besides, I think you actually wanted to make sure your pattern can match full strings that may contain line breaks. That means, you need (?s), Pattern.DOTALL, modifier:
String reg = "(?is).*Bad Word.*";
Also, to match any whitespace, you may rely on \s:
String reg = "(?ix).*Bad\\sWord.*"; // To only match 1 whitespace
String reg = "(?ix).*Bad\\s+Word.*"; // To account for 1 or more whitespaces

Password Validation with Regex Java

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.

Help with regex

I'm constructing a regex which will accept at least 1 alpha numerical character and any number of spaces.
Right now I've got...[A-Za-z0-9]+[ \t\r\n]* which I understand to be at least 1 alphanumeric OR at least 1 space. How would I fix this?
EDIT: To answer the comments below I want it to accept strings which contain ATLEAST 1 alphanumeric AND any number of (including no) spaces. Right now it will accept JUST a whitespace.
EDIT2: To clarify, I don't want the any number of whitespace (including 0) to be accepted unless there is at least 1 alphanumeric character

\s*\p{Alnum}[\p{Alnum}\s]*
Your regex, [A-Za-z0-9]+[ \t\r\n]*, requires the string to start with a letter or digit (or, more accurately, it doesn't start matching until it sees one). Adding \s* allows the match to start with whitespace, but you still won't match any alphanumerics after the first whitespace character that follows an alphanumeric (for example, it won't match the xyz in abc xyz. Changing the trailing \s* to [\p{Alnum}\s]* fixes that problem.
On a side note, \p{Alnum} is exactly equivalent to [A-Za-z0-9] in Java, which is not the case in all regex flavors. I used \p{Alnum}, not just because it's shorter, but because it gives more protection from typos like [A-z] (which is syntactically valid, but almost certainly not what the author really meant).
EDIT: Performance should be considered, too. I originally included a + after the first \p{Alnum}, but I realized that wasn't a good idea. If this were part of a longer regex, and the regex didn't match right away, it could end up wasting a lot of time trying to match the same groups of characters with \p{Alnum}+ or [\p{Alnum}\s]*. The leading \s* is okay, though, because \s doesn't match any of the characters that \p{Alnum} matches.

Any one or more word char zero or more whitespace
\w+\s*

Hey try this ([^\s]+\s*) [^\s] means catch everything that is not white space, while \s* means that an white space is optional (if you really want at least one white space put + instead of )
Edit: sory mine catch everithing not only alphanumeric (put ([a-zA-Z0-9]+\s) for alphanumeric)

This should do the trick:
\s*\p{Alnum}+\s*
\p{Alnum} is an alphanumeric character: [\p{Alpha}\p{Digit}]
* says "zero or more times"
+ says "at least one" (not "or" as you seem to believe, or is written |)
| means "or"
\s is a whitespace character: [ \t\n\x0B\f\r]
EDIT: To answer the comments below I want it to accept strings which contain AT LEAST 1 alphanumeric AND any number of (including no) spaces.
The pattern I suggested requires at least one alpha numeric character.
EDIT2: To clarify, I don't want the any number of whitespace (including 0) to be accepted unless there is at least 1 alphanumeric character
The pattern I suggested will not accept only white space characters only.