I have following regex
(\{\w*\}\s*[^{}]+\s*)\{?
and I am testing it on this string
this {match} is cool{match} but {match} this one is more cool
currently I am able to capture 2 groups -> {match} is cool and {match} this one is more cool, so as you can see group but {match} is missing.
Reason for this is because last matched character is {, so in next matching turn he will skip {, and won't be able to match until new { occurrence.
Does anyone knows how to force to match middle group also?
Debugging: http://regex101.com/r/hM5xE6/2
You can probably just remove the \{? (and the \s* too); you also don't need the capturing parentheses:
\{\w*\}[^{}]+
Test it live on regex101.com.
If you want to enforce the match to end before a { or at the end of the string, you can use a positive lookahead assertion for that:
\{\w*\}[^{}]+(?=\{|$)
But you would only need that if you wanted to avoid a match completely if there are nested braces, like in {{match} whatever}, where the first regex would find {match} whatever.
You can use the following regular expression to start matching from last letter in previous match
\{\w*\}[^{}]+(?=\{|$)
You are including the next { in the regex (but not in the match), so it begins the next match on the character after, skipping the first { and not matching until you get to the second.
There's no need for lookaheads or anything like that.
If you remove the trailing check for \{?, you get all 3 matches (can also remove the } from the brackets and the last \s*):
(\{\w*\}\s*[^{]+)
(http://regex101.com/r/hM5xE6/7)
you can also use the following regex, depending on how specific you need to be with the capture:
(\{\w*\}[\w\s]*)
http://regex101.com/r/hM5xE6/5
(\{\w*\}\s*[^{}]+\s*)(?=\{|$)
Try this.Use lookahead for 0 width assertion.See demo.
http://regex101.com/r/qC9cH4/18
Related
I have the following string;
Start: 738392E, 6726376N
I extracted 738392 ok using (?<=.art\:\s)([0-9A-Z]*). This gave me a one group match allowing me to extract it as a column value
.
I want to extract 6726376 the same way. Have only one group appear because I am parsing that to a column value.
Not sure why is (?=(art\:\s\s*))(?=[,])*(.*[0-9]*) giving me the entire line after S.
Helping me get it right with an explanation will go along way.
Because you used positive lookaheads. Those just make some assertions, but don't "move the head along".
(?=(art\:\s\s*)) makes sure you're before "art: ...". The next thing is another positive lookahead that you quantify with a star to make it optional. Finally you match anything, so you get the rest of the line in your capture group.
I propose a simpler regex:
(?<=(art\:\s))(\d+)\D+(\d+)
Demo
First we make a positive lookback that makes sure we're after "art: ", then we match two numbers, seperated by non-numbers.
There is no need for you to make it this complicated. Just use something like
Start: (\d+)E, (\d+)N
or
\b\d+(?=[EN]\b)
if you need to match each bit separately.
Your expression (?=(art\:\s\s*))(?=[,])*(.*[0-9]*) has several problems besides the ones already mentioned: 1) your first and second lookahead match at different locations, 2) your second lookahead is quantified, which, in 25 years, I have never seen someone do, so kudos. ;), 3) your capturing group matches about anything, including any line or the empty string.
You match the whole part after it because you use .* which will match until the end of the line.
Note that this part [0-9]* at the end of the pattern does not match because it is optional and the preceding .* already matches until the end of the string.
You could get the match without any lookarounds:
(art:\s)(\d+)[^,]+,\s(\d+)
Regex demo
If you want the matches only, you could make use of the PyPi regex module
(?<=\bStart:(?:\s+\d+[A-Z],)* )\d+(?=[A-Z])
Regex demo (For example only, using a different engine) | Python demo
I am having a problem parsing some fields from the following regular expression which I uploaded to rubular. The string that I am parsing is a special header from the banner of an FTP server. In order for me to process this banner, the line
special:pTXT1TOCAPTURE^:mTXT2TOCAPTURE^:uTXT3TOCAPTURE^
I thought that: (?i)^special(:[pmu](.*?)\^)?* would do the trick, however unfortunately this only gives me the last match and I am not sure why as I am lazily trying to capture each group. Also note that I should be able to capture an empty string also, i.e. if for ex the match string contains :u^
Wrap words Show invisibles Ruby version
Match result:
special:pTXT1TOMATCH^:mTXT2TOMATCH^:uTXT3TOMATCH^
Match groups:
:uTXT3TOMATCH^
TXT3TOMATCH
The idea is that the line must start with the test 'special' followed by up to 3 capture groups delimited with p,m or u lazily up to the next ^ symbol. I need to capture the text indicated above - basically I need to find TXT1TOCAPTURE, TXT2TOCAPTURE, and TXT3TOCAPTURE. There should be at least one of these three capture groups.
Thanks in advance
You have two problems with your RegEx, one is syntactic and one is conceptual.
Syntactic:
We don't have such a modifier ?* in PCRE but it is equal to * in Ruby which denotes a greedy quantifier. In the case of applying to a capturing group it captures last match.
Conceptual:
Using a lazy quantifier .*? doesn't provide you with continues matches. It stops immediately on engine satisfaction. While g modifier is on next match will never occur as there is no ^special at the next position of last match.
Solution is using \G token to benefit from its mean of start matching at the end of previous match:
(?:special|(?!\A)\G):([pmu][^^]*\^)
Live demo
You might want to have the \G modifier:
(?:(?:^special:)|\G(?!\A)\^:)[pmu]([^^]+)
See it working on rubular.com.
I have regex that I am trying to match to specific function parameters. I want to be able to style them a certain way in a language package.
Here is the text I am trying to match:
addFill(path:svgjs.Element, pattern:Pattern, docMaxSide:number) {
pathFillId(path)
}
In this example, I want to match the words "path" "pattern" and "docMaxSide" from the parameters. I want to make sure it does NOT match the word "path" in the second line (where I am calling pathFillId).
Here is my current regex: \(.*?(\w+):.*?\)
Broken down:
\( Find open parens
.*? It may have stuff before it, but after the parens
(\w+): Capture a word before a colon
.*? There may be more stuff after the colon
\) Close parens
Right now, it will only match the first item, "path". But I need it to match all the words I mentioned above.
UPDATE: I should have been more specific. It should only match if it's a function parameter. For example, I don't want path1 matched in the following: var path1:string. The difficulty is coming up with regex that matches items only between parens.
Try this:
\w+(?=:)
with the g modifier (the global modifier finds all elements and don't return on the first match)
Also see the example
UPDATE
If you want only match the parameters in the parenthesis you can do this:
\w+(?=:[\w.]+\s*[,)])
Here is the example for this regex
You problem is this part of your regex: .*?. So you specify that you want any character (.), that's correct. But then you must decide for one of * and ? - * means {0,}, ? means {0,1}.
If that doesn't help, you might test your regex with regexe.com or similar.
I've researched around for a while and haven't found a clue for matching the following pattern (I am also very new to regex, though), it looks either like
/abc/foo/bar(/*)
or
/abc/foo/bar/stop
So I want to match and capture the above string as /abc/foo/bar. Now "/stop" is an optional string that could be appended at the end of the pattern. The goal is to get the desired capture while ignoring "stop" if they present (and if "stop" exists multiple times stop at the first "stop"), while allow as many slashes in the middle as possible except the slash at the end of line.
If I simply do:
^(/.*[^/])/*$
Which is greedy in including all slashes until I strip off the possible last occurrence; but in order to accept the second case where I have an optional "/stop", I need to match in a non-greedy way until I find the first possible "/stop" and stop there.
How can I craft a single regex that matches both cases?
EDIT: Not sure if my previous example wasn't clear enough. Try to give more, say I want to match and capture "/abc/foo/bar" in all of the following strings:
/abc/foo/bar
/abc/foo/bar/
/abc/foo/bar///
/abc/foo/bar/stop
/abc/foo/bar/stop/foo/bar/stop/stop
/abc/foo/bar//stop
While it won't match any of the followings:
/abc/foo/bar/sto (will match the whole "/abc/foo/bar/sto" instead)
/abc/foo/bar/abc/foo/bar (it will catch "/abc/foo/bar/abc/foo/bar" instead)
Let me know if this is clear enough. Thanks!
Try this:
/^(?:\/+(?!$|(?:stop\/?))[^\/]+)*/
Regex101 Demo
Explanation:
This matches the start of the string (^), followed by zero or more instances of the following pattern:
one or more slashes (\/+) that are not followed by the end of the string ($) or by stop, followed by
one or more non-slash characters ([^\/]+)
Here's a Debuggex Demo with working unit tests.
EDIT: Here is an alternative, arguably simpler, regex:
/^.+?(?=\/*$|\/+stop\b)/
This matches one or more characters in a non-greedy manner, then stops when whatever is after the match is one of the following:
the end of the string ($), possibly preceded by one or more slashes (\/*)
one or more slashes, the word stop, and a word break.
Here's a Regex101 demo of this option.
EDIT 2: If you'd like to test this, here's a simple JavaScript test that tests the second regex above against various test strings and logs the results to the console:
var re = /^.+?(?=\/*$|\/+stop\b)/,
test_strings = ["/abc/foo/bar",
"/abc/foo/bar/",
"/abc/foo/bar///",
"/abc/foo/bar/stop",
"/abc/foo/bar/stop/foo/bar/stop/stop",
"/abc/foo/bar//stop",
"/abc/foo/bar/sto",
"/abc/foo/bar/abc/foo/bar"];
for(var s = 0; s < test_strings.length; s++) {
console.log(test_strings[s].match(re)[0]);
}
/*
Results:
/abc/foo/bar
/abc/foo/bar
/abc/foo/bar
/abc/foo/bar
/abc/foo/bar
/abc/foo/bar
/abc/foo/bar/sto
/abc/foo/bar/abc/foo/bar
*/
You can try something like this:
^((?:/[^/]+)+?)(?:/+|/+stop(?:/.*)?)$
demo
and if atomic groups are available, you better write:
^((?:/[^/]+)+?)(?>/+$|/+stop(?:/.*)?)
demo
If lookaheads are available:
^/(?>[^/]+|/(?!/*(?:$|stop(?:/|$))))+
demo
ps: don't forget to escape slashes if your delimiters are slashes.
As Ed Cottrell notices it, features like atomic grouping are not available in language like Javascript or in the re module of Python. However, this feature can be efficiently emulated using the fact that a lookahead is naturaly atomic: (?>a+) <=> (?=(a+))\1
Hey guys I've been working with this one for a little while. I can't seem to get it.
Here is what I have so far
(#[^{2,}+)([^(\s\W\d{2}]+)(\b)
http://rubular.com/r/zlx3j00Wjl
Although this is not excepting periods in the match.
I basically need to match this.
#function.name(param)
I just need to match function.name. This does that.
http://rubular.com/r/hWMB72LsWT
I don't want to match this
##function.name(param)
hello##test.com`
Didn't know if anyone has any ideas. Thanks for the help.
You can use a negative lookahead: #(?!#) matches a # not followed by another #.
Here is my go at it (here it is on Rubular):
(?<!#)#(\w+(?:\.\w+)*)\([^)]*\)
Explained:
(?<!#)# # an '#' not preceded by an '#'
(\w+(?:\.\w+)*) # any number of xxx.xxx.xxx, captured into a group
\([^)]*\) # brackets, containing anything that isn't a closing bracket
Since this is Ruby, you might not care about matching parentheses. In that case you can just remove the last section.
Try this:
(?:^|\s)#+([^(]+)
You will have function.match and function.name in the first group, will not match hello##test.com. Rubular:
http://rubular.com/r/b8gy1LcVGz
Try this
(?!.*##)^#([^()\s]+)\b
See it here on Rubular
I removed some brackets from your expression
I removed the Quantifier from the leading #
(?!.*##) is a negative lookahead assertion. It will fail if it finds anywhere in the string two # characters in a row.
I am not sure about your requirements, if there is all the time a set of brackets at the end, then you don't need your word boundary. If there can be similar strings without brackets that you don't want to match, then I would add another lookahead to ensure this assertion:
?!.*##)^#([^()\s]+)(?=\()
See it here on Rubular