I'm trying to extract a text after a sequence. But I have multiple sequences. the regex should ideally match first occurrence of any of these sequences.
my sequences are
PIN, PIN :, PIN IN, PIN IN:, PIN OUT,PIN OUT :
So I came up with the below regex
(PIN)(\sOUT|\sIN)?\:?\s*
It is doing the job except that the regex is also matching strings like
quote lupin in, pippin etc.
My question is how can I strictly select the string that match the pattern being the whole word
note: I tried ^(PIN)(\sOUT|\sON)?\:?\s* but of no use.
I'm new to java, any help is appreciated
It’s always recommended to have the documentation at hand when using regular expressions.
There, under Boundary matchers we find:
\b A word boundary
So you may use the pattern \bPIN(\sOUT|\sIN)?:?\s* to enforce that PIN matches at the beginning of a word only, i.e. stands at the beginning of a string/line or is preceded by non-word characters like space or punctuation. A boundary only matches a position, rather than characters, so if a preceding non-word character makes this a word boundary, the character still is not part of the match.
Note that the first (…) grouping was unnecessary for the literal match PIN, further the colon : has no special meaning and doesn’t need to be escaped.
Related
I'm not familiar yet with java regular expressions. I want to validate a string that has the following format:
String INPUT = "[name1 value1];[name2 value2];[name3 value3];";
namei and valuei are Strings should contain any characters expect white-space.
I tried with this expression:
String REGEX = "([\\S*\\s\\S*];)*";
But if I call matches() I get always false even for a good String.
what's the best regular expression for it?
This does the trick:
(?:\[\w.*?\s\w.*?\];)*
If you want to only match three of these, replace the * at the end with {3}.
Explanation:
(?:: Start of non-capturing group
\[: Escapes the [ sign which is a meta-character in regex. This
allows it to be used for matching.
\w.*?: Lazily matches any word character [a-z][A-Z][0-9]_. Lazy matching means it attempts to match the character as few times possible, in this case meaning that when will stop matching once it finds the following \s.
\s: Matches one whitespace
\]: See \[
;: Matches one semicolon
): End of non-capturing group
*: Matches any number of what is contained in the preceding non-capturing group.
See this link for demonstration
You should escape square brackets. Also, if your aim is to match only three, replace * with {3}
(\[\\S*\\s\\S*\];){3}
Using Java, i want to detect if a line starts with words and separator then "myword", but this regex takes too long. What is incorrect ?
^\s*(\w+(\s|/|&|-)*)*myword
The pattern ^\s*(\w+(\s|/|&|-)*)*myword is not efficient due to the nested quantifier. \w+ requires at least one word character and (\s|/|&|-)* can match zero or more of some characters. When the * is applied to the group and the input string has no separators in between word characters, the expression becomes similar to a (\w+)* pattern that is a classical catastrophical backtracking issue pattern.
Just a small illustration of \w+ and (\w+)* performance:
\w+: (\w+)*
You pattern is even more complicated and invloves more those backtracking steps. To avoid such issues, a pattern should not have optional subpatterns inside quantified groups. That is, create a group with obligatory subpatterns and apply the necessary quantifier to the group.
In this case, you can unroll the group you have as
String rx = "^\\s*(\\w+(?:[\\s/&-]+\\w+)*)[\\s/&-]+myword";
See IDEONE demo
Here, (\w+(\s|/|&|-)*)* is unrolled as (\w+(?:[\s/&-]+\w+)*) (I kept the outer parentheses to produce a capture group #1, you may remove these brackets if you are not interested in them). \w+ matches one or more word characters (so, it is an obligatory subpatter), and the (?:[\s/&-]+\w+)* subpattern matches zero or more (*, thus, this whole group is optional) sequences of one or more characters from the defined character class [\s/&-]+ (so, it is obligatory) followed with one or more word characters \w+.
I'm not sure if this is possible using Regex but I'd like to be able to limit the number of underscores allowed based on a different character. This is to limit crazy wildcard queries to a search engine written in Java.
The starting characters would be alphanumeric. But I basically want a match if there are more underscores than preceding characters. So
BA_ would be fine but BA___ would match the regex and would get kicked out of the query parser.
Is that possible using Regex?
Yes you can do it. This pattern will succeed only if there are less underscores than letters (you can adapt it with the characters you want):
^(?:[A-Z](?=[A-Z]*(\\1?+_)))*+[A-Z]+\\1?$
(as Pshemo notices it, anchors are not needed if you use the matches() method, I wrote them to illustrate the fact that this pattern must be bounded whatever the means. With lookarounds for example.)
negated version:
^(?:[A-Z](?=[A-Z]*(\\1?+_)))*\\1?_*$
The idea is to repeat a capture group that contains a backreference to itself + an underscore. At each repetition, the capture group is growing. ^(?:[A-Z](?=[A-Z]*+(\\1?+_)))*+ will match all letters that have a correspondant underscore. You only need to add [A-Z]+ to be sure that there is more letters, and to finish your pattern with \\1? that contains all the underscores (I make it optional, in case there is no underscore at all).
Note that if you replace [A-Z]+ with [A-Z]{n} in the first pattern, you can set exactly the number of characters difference between letters and underscores.
To give a better idea, I will try to describe step by step how it works with the string ABC-- (since it's impossible to put underscores in bold, i use hyphens instead) :
In the non-capturing group, the first letter is found
ABC-- ^(?:[A-Z](?=[A-Z]*(\1?+-)))*+[A-Z]+\1?$
let's enter the lookahead (keep in mind that all in the lookahead is only
a check and not a part of the match result.)
ABC-- ^(?:[A-Z](?=[A-Z]*(\1?+-)))*+[A-Z]+\1?$
ABC-- ^(?:[A-Z](?=[A-Z]*(\1?+-)))*+[A-Z]+\1?$
the first capturing group is encounter for the first time and its content is not
defined. This is the reason why an optional quantifier is used, to avoid to make
the lookahead fail. Consequence: \1?+ doesn't match something new.
ABC-- ^(?:[A-Z](?=[A-Z]*(\1?+-)))*+[A-Z]+\1?$
the first hyphen is matched. Once the capture group closed, the first capture
group is now defined and contains one hyphen.
ABC-- ^(?:[A-Z](?=[A-Z]*(\1?+-)))*+[A-Z]+\1?$
The lookahead succeeds, let's repeat the non-capturing group.
ABC-- ^(?:[A-Z](?=[A-Z]*(\1?+-)))*+[A-Z]+\1?$
The second letter is found
ABC-- ^(?:[A-Z](?=[A-Z]*(\1?+-)))*+[A-Z]+\1?$
We enter the lookahead
ABC-- ^(?:[A-Z](?=[A-Z]*(\1?+-)))*+[A-Z]+\1?$
but now, things are different. The capture group was defined before and
contains an hyphen, this is why \1?+ will match the first hyphen.
ABC-- ^(?:[A-Z](?=[A-Z]*(\1?+-)))*+[A-Z]+\1?$
the literal hyphen matches the second hyphen in the string. And now the
capture group 1 contains the two hypens. The lookahead succeeds.
ABC-- ^(?:[A-Z](?=[A-Z]*(\1?+-)))*+[A-Z]+\1?$
We repeat one more time the non capturing group.
ABC-- ^(?:[A-Z](?=[A-Z]*(\1?+-)))*+[A-Z]+\1?$
In the lookahead. There is no more letters, it's not a problem, since
the * quantifier is used.
ABC-- ^(?:[A-Z](?=[A-Z]*(\1?+-)))*+[A-Z]+\1?$
\\1?+ matches now two hyphens.
ABC-- ^(?:[A-Z](?=[A-Z]*(\1?+-)))*+[A-Z]+\1?$
but there is no more hyphen in the string for the literal hypen and the regex
engine can not use the bactracking since \1?+ has a possessive quantifier.
The lookahead fails. Thus the third repetition of the non-capturing group too!
ABC-- ^(?:[A-Z](?=[A-Z]*(\1?+-)))*+[A-Z]+\1?$
ensure that there is at least one more letter.
ABC-- ^(?:[A-Z](?=[A-Z]*(\1?+-)))*+[A-Z]+\1?$
We match the end of the string with the backreference to capture group 1 that
contains the two hyphens. Note that the fact that this backreference is optional
allows the string to not have hyphens at all.
ABC-- ^(?:[A-Z](?=[A-Z]*(\1?+-)))*+[A-Z]+\1?$
This is the end of the string. The pattern succeeds.
ABC-- ^(?:[A-Z](?=[A-Z]*(\1?+-)))*+[A-Z]+\1?$
Note: The use of the possessive quantifier for the non-capturing group is needed to avoid false results. (Where you can observe a strange behavior, that can be useful.)
Example:ABC--- and the pattern: ^(?:[A-Z](?=[A-Z]*(\1?+-)))*[A-Z]+\1?$ (without the possessive quantifier)
The non-capturing group is repeated three times and `ABC` are matched:
ABC--- ^(?:[A-Z](?=[A-Z]*(\1?+-)))*[A-Z]+\1?$
Note that at this step the first capturing group contains ---
But after the non capturing group, there is no more letter to match for [A-Z]+
and the regex engine must backtrack.
ABC--- ^(?:[A-Z](?=[A-Z]*(\1?+-)))*[A-Z]+\1?$
Question: How many hyphens are in the capture group now?
Answer: Always three!
If the repeated non-capturing group gives a letter back, the capture group contains always three hyphens (as the last time the capture group has been read by the regex engine).This is counter-intuitive, but logical.
Then the letter C is found:
ABC--- ^(?:[A-Z](?=[A-Z]*(\1?+-)))*[A-Z]+\1?$
And the three hyphens
ABC--- ^(?:[A-Z](?=[A-Z]*(\1?+-)))*[A-Z]+\1?$
The pattern succeeds
ABC--- ^(?:[A-Z](?=[A-Z]*(\1?+-)))*[A-Z]+\1?$
Robby Pond asked me in comments how to find strings that have more underscores than letters (all that is not an underscore). The best way is obviously to count the numbers of underscores and to compare with the string length. But about a full regex solution, it is not possible to build a pattern for that with Java since the pattern needs to use the recursion feature. For example you can do it with PHP:
$pattern = <<<'EOD'
~
(?(DEFINE)
(?<neutral> (?: _ \g<neutral>?+ [A-Z] | [A-Z] \g<neutral>?+ _ )+ )
)
\A (?: \g<neutral> | _ )+ \z
~x
EOD;
var_dump(preg_match($pattern, '____ABC_DEF___'));
Its not possible in singular regular expression.
i) Logic needs to be implemented to get number of characters before underscores(regular expression should be written to get characters word before underscore).
ii) And validate result (number of characters - 1) = number of semicolons followed(regular expression which returns stream of underscores followed by characters).
Edit: Dang! I just noticed that you need this for java. Anyways...I leave it here if someone from the .Net world stumbles upon this post.
You can use Balancing Groups if you are using .Net:
^(?:(?<letter>[^_])|(?<-letter>_))*(?(letter)(?=)|(?!))$
The .net regex engine has the ability to maintain all captured patterns in the captured groups. In other flavors the captured group would always contain the last matched pattern but in .net all previous matches are contained in a capture collection for your use. Also the .net engine has the ability to push and pop to the stack of the captured groups using the ?<group-name>, ?<-group-name> constructs. These two handy constructs can be utilized to match pairs of paranthesis, etc.
In the above regex, the engine starts from the start of the string and tries to match anything other than "_". This of course can be changed to whatever works for you(e.g [A-Z][a-z]). The alternation basically means either match [^\_] or [\_] and doing so either push or pop from the captured group.
The latter part of the regex is a conditional (?(group-name)true|false). It basically says, if the group still exists(more pushes than pops), then do the true section and if not do the false section. The easiest way to make the pattern match is to use an empty positive look ahead: (?=) and the easiest way to make it fail is (?!) which is a negative lookahead.
I am trying to match everything but garbage values in the entire string.The pattern I am trying to use is:
^.*(?!\w|\s|-|\.|[#:,]).*$
I have been testing the pattern on regexPlanet and this seems to be matching the entire string.The input string I was using was:
Vamsi///#k03#g!!!l.com 123**5
How can I get it to only match everything but the pattern,I would like to replace any string that matches with an empty space or a special charecter of my choice.
The pattern, as written, is supposed to match the whole string.
^ - start of string.
.* - zero or more of any character.
(?!\w|\s|-|\.|[#:,]) - negative look-ahead for some characters.
.* - zero or more of any character.
$ - end of string.
If you only want to match characters which aren't one of the supplied characters, try simply:
[^-\w\s.#:,]
[^...] is a negated character class, it will match any characters not supplied in the brackets. See this for more information.
Test.
I have a regular expression,
end\\s+[a-zA-Z]{1}[a-zA-Z_0-9]
which is supposed to match a line with the specifications
end abcdef123
where abcdef123 must start with a letter and subsequent alphanumeric characters.
However currently it is also matching this
foobar barfooend
bar fred bob
It's picking up that end at the end of barfooend and also picking up bar in effect returning end bar as a legitimate result.
I tried
^end\\s+[a-zA-Z]{1}[a-zA-Z_0-9]
but that doesn't seem to work at all. It ends up matching nothing.
It should be fairly simple but I can't seem to nut it out.
\s includes also newline characters. So you either need to specify a character class that has only the wanted whitespace charaters or exclude the not wanted.
Use instead of \\s+ one of those:
[^\\S\r\n] this includes all whitespace but not \r and \n. See end[^\S\r\n]+[a-zA-Z][a-zA-Z_0-9]+ here on Regexr
[ \t] this includes only space and tab. See end[ \t]+[a-zA-Z][a-zA-Z_0-9]+ here on Regexr
You can use \b (word boundary detection) to check a word boundary. In our case we will use it to match the beginning of the word end. It can also be used to match the end of a word.
As #nhahtdh stated in his comment the {1} is redundant as [a-zA-Z] already matches one letter in the given range.
Also your regex does not do what you want because it only matches one alphanumeric character after the first letter. Add a + at the end (for one or more times) or * (for zero or more times).
This should work:
"\\bend\\s+[a-zA-Z]{1}[a-zA-Z_0-9]*"
Edit : I think \b is better than ^ because the latter only matches the beginning of a line.
For example take this input : "end azd123 end bfg456" There will be only one match for ^ when \b will help matching both.
Try the regular expression:
end[ ]+[a-zA-Z]\w+
\w is a word character: [a-zA-Z_0-9]