I'm very lost with a regular expression. It's just black magic to me. Here's what i need:
there is a filename: some_file.jpg
it might be in the following format: some_file_p250.jpg
the regex to match the file in simple format: /^([a-zA-Z_-0-9]+).(jpg|jpeg|png)$/
the regex to match the file in advanced format: /^([a-zA-Z_-0-9]+)(_[a-z]?[0-9]{2,3}).(jpg|jpeg|png)$/
my question is as follows: how do i make the "(_[a-z]?[0-9]{3,4})" part optional? I've tried adding a question mark to the second group like this:
/^([a-zA-Z_\-0-9]+)(_[a-z]?[0-9]{3,4})?\.(jpg|jpeg|png)$/
Even though the pattern works, it always captures the contents of the second group in the first group and leaves the second empty.
How can i make this work to capture the filename, advanced part (_p250) and the extension separately? I'm thinking it has something to do with the greediness of the first group, but i might be completely wrong and even if i'm right, i still don't know how to solve it.
Thanks for your thoughts
Adding a question mark after the first plus will make the first capturing expression non-greedy. This worked for me using your test case:
/^([a-zA-Z_\-0-9]+?)(_[a-z]?[0-9]{3,4})?\.(jpg|jpeg|png)$/
I tested in Javascript, not PHP, but here's my test:
"some_file_p250.jpg".match(/^([a-zA-Z_\-0-9]+?)(_[a-z]?[0-9]{3,4})?\.(jpg|jpeg|png)$/)
and my results:
["some_file_p250.jpg", "some_file", "_p250", "jpg"]
In my experience, making a capturing expression non-greedy makes regular expressions a lot more intuitive and will often make them work the way I expect them to work. In your case, it was doing what you suspected; the first expression was capturing everything and never gave the second expression a chance to capture anything.
I think this is what you want:
/^([a-zA-Z_\-0-9]+)(|_[a-z]?[0-9]{3,4})?\.(jpg|jpeg|png)$/
or
/^([\d\w\-]+)(|_[a-z]?[0-9]{3,4})\.(jpg|jpeg|png)$/
Related
In Rust with the Regex crate, I've been trying to wrap my head around a regex expression to capture and extract things between square brackets [] yet exclude the brackets from the capture. Given:
// template[tags(foo,bar,baz)]
# template[replace_all(foo:bar)]
I'd like:
tags(foo,bar,baz)
replace_all(foo:bar)
I can easily get the [] capture group but i'm not understanding how to capture with an exclusion of characters after the match. I've been manually replacing these but it seems gross to me. I would love to be able to do it all in one expression.
Update: I am aware that I can get these in multiple capture groups but i'm really curious if there's a way to only capture the single one - hence exclude.
Looking over the docs i'm just not pickin up a way this can be done. There's a lot of great examples using look aheads and behinds but that doesn't appear to be apart of the rust regex crate. Am i missing something obvious here? Thanks for the help.
I am trying to write a regular expression that will match values such as U.S., D.C., U.S.A., etc.
Here is what I have so far -
\b([a-zA-Z]\.){2,}+
Note how this expression matches but does not include the last letter in the acronym.
Can anyone help explain what I am missing here?
SOLUTION
I'm posting the solution here in case this helps anyone.
\b(?:[a-zA-Z]\.){2,}
It seems as if a non-capturing group is required here.
Try (?:[a-zA-Z]\.){2,}
?: (non-capturing group) is there because you want to omit capturing the last iteration of the repeated group.
For example, without ?:, 'U.S.A.' will yield a group match 'A.', which you are not interested about.
None of these proposed solutions do what yours does - make sure that there are at least 2 letters in the acronym. Also, yours works on http://rubular.com/ . This is probably some issue with the regex implementation - to be fair, all of the matches that you got were valid acronyms. To fix this, you could either:
Make sure there's a space or EOF succeeding your expression ((?=\s|$) in ruby at least)
Surround your regex with ^ and $ to make sure it catches the whole string. You'd have to split the whole string on spaces to get matches with this though.
I prefer the former solution - to do this you'd have:
\b([a-zA-Z]\.){2,}(?=\s|$)
Edit: I've realized this doesn't actually work with other punctuation in the string, and a couple of other edge cases. This is super ugly, but I think it should be good enough:
(?<=\s|^)((?:[a-zA-Z]\.){2,})(?=[[:punct:]]?(?:\s|$))
This assumes that you've got this [[:punct:]] character class, and allows for 0-1 punctuation marks after an acronym that won't be captured. I've also fixed it up so that there's a single capture group that gets the whole acronym. Check out validation at http://rubular.com/r/lmr0qERLDh
Bonus: you now get to make this super confusing to anyone reading it.
This should work:
/([a-zA-Z]\.)+/g
I have slightly modified the solution above:
\b(?:[a-zA-Z]+\.){2,}
to enable capturing acronyms containing more than one letter between the dots, like in 'GHQ.AFP.X.Y'
I had a regex expression
^\d{9}_[a-zA-Z]{1}_(0[1-9]|1[0-2]).(0[1-9]|[1-2][0-9]|3[0-1]).[0-9]{4}_\d*_[0-9a-zA-Z]*_[0-9a-zA-Z]*
and string that match regex expression
000066874_A_12.31.2014_001_2Q_ICAN14
if user by mistake enters the string other than above format like
000066874_12.31.14_001_2Q_ICAN14
I need to find out in which part of my regex got failed. I tried using Regex.Matches and Regex.Match but using this I couldn't find in which part my string got miss matched with my Regex expression. I am using vb.net
This is very complicated to do with regex. I managed to make this regex, but you still have to check the capture groups after that.
^(?:(?:(\d{9})|.*?)_)?(?:(?:([a-zA-Z]{1})|.*?)_)?(?:(?:((?:0[1-9]|1[0-2]).(?:0[1-9]|[1-2][0-9]|3[0-1]).[0-9]{4})|.*?)_)?(?:(?:(\d*)|.*?)_)?(?:(?:([0-9a-zA-Z]*)|.*?)_)?(?:([0-9a-zA-Z]*)|.*?)$ will work if you, as seen in demo: https://regex101.com/r/aJ1wG1/2
Each part before an underline is a capture group, if a capture group is not there, there's an error in it. As you can see in the example, $3 is not present in 1st example, hence, a mistake in date is there. In second example, the $2 is not present, hence $2 onward are not there. 3rd example is correct and all 6 caputre groups are there.
When regexes get this massive, it's a sign that probably a different method should be used to solve the problem, but this might work for you with some additional code for group result checks.
I currently have this regex: this\.(.*)?\s[=,]\s, however I have come across a pickle I cannot fix.
I tried the following Regex, which works, but it captures the space as well which I don't want: this\.(.*)?(?<=\s)=|(?<!\s),. What I'm trying to do is match identifier names. An example of what I want and the result is this:
this.""W = blah; which would match ""W. The second regex above does this almost perfectly, however it also captures the space before the = in the first group. Can someone point me in the correct direction to fix this?
EDIT: The reason for not simply using [^\s] in the wildcard group is that sometimes I can get lines like this: this. "$ = blah;
EDIT2: Now I have another issue. Its not matching lines like param1.readBytes(this.=!3,0,param1.readInt()); properly. Instead of matching =!3 its matching =!3,0. Is there a way to fix this? Again, I cannot simply use a [^,] because there could be a name like param1.readBytes(this.,3$,0,param1.readInt()); which should match ,3$.
(.*) will match any character including whitespace.
To force it not to end in whitespace change it to (.*[^\s])
Eg:
this\.(.*[^\s])?\s?[=,]\s
For your second edit, it seems like you are doing a language parser. Even though regular expressions are powerful, they do have limits. You need a grammar parser for that.
Maybe you can tell in your first block to capture non space characters, instead of any.
this\.(\S*)?(?<=\s)=|(?<!\s),
Sorry for the confusing title, I could not think of the correct wording. I am trying to understand if there is a way for regex to match different strings, depending on whether a previous capture group was captured or not.
/th?u(e|r)sday/
This matches tuesday, thursday but also thuesday and tursday. Is there any way to indicate in the regex that a part should only match, if a previous part was matched... so I imagine a potential syntax like... (?#:pattern) where # is a number from a capturing group, and if the capturing group captured, then pattern is included, otherwise it is skipped. A similar pattern (!#:pattern) for if the #th group is not captured. This invented syntax is to demonstrate what I am trying to do. With this invented syntax, I could solve my problem above like this...
/t(h)?u(!1:e)(?1:r)sday/
Is there any such syntax in regex to achieve this type of referencing?
This feature does exist in some regex implementations, and the regex from your example would be written like this:
/t(h)?u(?(1)r|e)sday/
Obviously this is not the best example, since /t(hur|ue)sday/ is equivalent and much shorter, but there are cases where this is more useful.
Check out the second to last element in the table of this advanced regex reference page, with additional information on conditionals here.
Syntax:
(?(1)then|else)
Description:
If the first capturing group took part in the match attempt thus far, the "then" part must match for the overall regex to match. If the first capturing group did not take part in the match, the "else" part must match for the overall regex to match.
Example:
(a)?(?(1)b|c) matches ab, the first c and the second c in babxcac
According to the same page, conditionals are supported by the JGsoft engine, Perl, PCRE and the .NET framework.
Why not just use a more specific disjunction?
/t(hur|ue)sday/