I'm really bad at regex and still learning. I'm trying to setup my regex to find my first URI below.
/test/guid/5824812d100afbc60ef09411
/test/guid/5824812d100afbc60ef09411/action/create
/test/guid/5824812d100afbc60ef09411/action/version/delete
I have my regex working for both the (/action/create) and (/action/version/delete).
I need the first to be it's own individual URI. The guid after /guid changes, but it never will contain anything after.
These are working:
\/test\/guid\/\d.*\/action\/create
\/test\/guid\/\d.*\/action\/version\/delete
However if I use the same convention to find the first URI, it finds them all. I need all 3 separate.
Help?
Anchors are your friend here. ^ matches the beginning of a line (or beginning of the full string, depending on your modifiers) and $ matches the end.
So all you need is something like this:
\/test\/guid\/[a-z0-9]+$
That should be good enough, since after the guid's string of alphanumeric characters you're expecting the string to either terminate or have a forward slash, but if your guid is of a known fixed length, it might be better to do something like:
\/test\/guid\/[a-z0-9]{24}$
Related
I've got a practical application for a vim regex where I'd like to remove numbers from the end of file location links. For example, if the developer is sloppy and just adds files and doesn't reuse file locations, you'll end up with something awful like this:
PATH_TO_MY_FILES>
PATH_TO_MY_FILES1>
...
PATH_TO_MY_FILES22>
PATH_TO_MY_FILES_ELSEWHERE>
PATH_TO_MY_FILES_ELSEWHERE1>
...
So all I want to do is to S&R and replace PATH_TO_MY_FILES*\d+ with PATH_TO_MY_FILES* using regex. Obviously I am not doing it quite right, so I was hoping someone here could not spoon feed the answer necessarily, but throw a regex buzzword my way to get me on track.
Here's what I have tried:
:%s\(PATH_TO_MY_FILES\w*\)\(\d+\)>:gc
But this doesn't work, i.e. if I just do a vim search on that, it doesn't find anything. However, if I use this:
:%s\(PATH_TO_MY_FILES\w*\)\(\d\)>:gc
It will match the string, but the grouping is off, as expected. For example, the string PATH_TO_MY_FILES22 will be grouped as (PATH_TO_MY_FILES2)(2), presumably because the \d only matches the 2, and the \w match includes the first 2.
Question 1: Why doesn't \d+ work?
If I go ahead and use the second string (which is wrong), Vim appears to find a match (even though the grouping is wrong), but then does the replacement incorrectly.
For example, given that we know the \d will only match the last number in the string, I would expect PATH_TO_MY_FILES22> to get replaced with PATH_TO_MY_FILES2>. However, instead it replaces it with this:
PATH_TO_MY_FILES2PATH_TO_MY_FILES22>gt
So basically, it looks like it finds PATH_TO_MY_FILES22>, but then replaces only the & with group 1, which is PATH_TO_MY_FILES2.
I tried another regex at Regexr.com to see how it would interpret my grouping, and it looked correct, but maybe a hack around my lack of regex understanding:
(PATH_TO_\D*)(\d*)>
This correctly broke my target string into the PATH part and the entire number, so I was happy. But then when I used this in Vim, it found the match, but still replaced only the &.
Question 2: Why is Vim only replacing the &?
Answer 1:
You need to escape the + or it will be taken literally. For example \d\+ works correctly.
Answer 2:
An unescaped & in the replacement portion of a substitution means "the entire matched text". You need to escape it if you want a literal ampersand.
I'm trying to filter out links to videos from different video portals currently. The main problem I'm facing is that certain advertisement links got my video link in them, where regex tells me it "matches". I do not want that. Is there a way that I, for example, can say:
When there is anything before the http://, it won't match. How would it be for this example URL?
/http:\/\/example.com/
I tried something with ^[.] because I read the dot stands for any character, but it doesn't really work out.
You could change what does matches so that it's only the beginning of the string. for example:
/^http[s]?:\/\/example.com
Would only match if the string begins with http://example.com or https://example.com. However, this won't work if you're trying to scan for a match in the middle of a block of text.
This is where the anchor ^ comes to work:
/^http:\/\/example.com/
The caret ^ matches the position before the first character in the string. In another word, it means http must be the start of the string.
How can I successfully parse the text below in that format to parse just
To: User <test#test.com>
and
To: <test#test.com>
When I try to parse the text below with
/To:.*<[A-Z0-9._+-]+#[A-Z0-9.-]+\.[A-Z]{2,4}>/mi
It grabs
Message-ID <CC2E81A5.6B9%test#test.com>,
which I dont want in my answer.
I have tried using $ and \z and neither work. What am I doing wrong?
Information to parse
To: User <test#test.com> Message-ID <CC2E81A5.6B9%test#test.com>
To:
<test#test.com>
This is my parsing information in Rubular http://rubular.com/r/DQMQC4TQLV
Since you haven't specified exactly what your tool/language is, assumptions must be made.
In general regex pattern matching tends to be aggressive, matching the longest possible pattern. Your pattern starts off with .*, which means that you're going to match the longest possible string that ENDS WITH the remainder of your pattern <[A-Z0-9._+-]+#[A-Z0-9.-]+\.[A-Z]{2,4}>, which was matched with <CC2E81A5.6B9%test#test.com> from the Message-ID.
Both Apalala's and nhahtdh's comments give you something to try. Avoid the all-inclusive .* at the start and use something that's a bit more specific: match leading spaces, or match anything EXCEPT the first part of what you're really interested in.
You need to make the wildcard match non greedy by adding a question mark after it:
To:.*?<[A-Z0-9._+-]+#[A-Z0-9.-]+\.[A-Z]{2,4}>
I have Googled it, and found the following results:
http://icfun.blogspot.com/2008/03/regular-expression-to-handle-negative.html
http://regexlib.com/DisplayPatterns.aspx?cattabindex=2&categoryId=3
With some (very basic) Regex knowledge, I figured this would work:
r\.(^-?\d+)\.(^-?\d+)\.mcr
For parsing such strings:
r.0.0.mcr
r.-1.5.mcr
r.20.-1.mcr
r.-1.-1.mcr
But I don't get a match on these.
Since I'm learning (or trying to learn) Regex, could you please explain why my pattern doesn't match (instead of just writing a new working one for me)? From what I understood, it goes like so:
Match r
Match a period
Match a prefix negative sign or not, and store the group
Match a period
Match a prefix negative sign or not, and store the group
Match a preiod
Match mcr
But I'm wrong, apparently :).
You are very close. ^ matches the start of a string, so it should only be located at the start of a pattern (if you want to use it at all - that depends on whether you will also accept e.g. abcr.0.0.mcr or not). Similarly, one can use $ (but only at the end of the pattern) to indicate that you will only accept strings that do not contain anything after what the pattern matches (so that e.g. r.0.0.mcrabc won't be accepted). Otherwise, I think it looks good.
The ^ characters are telling it to match only at the beginning of a line; since it's obviously not at the beginning of a line in either case, it fails to match. In this case, you just need to remove both ^s. (I think what you're trying to say is "don't let anything else be in between these", but that's the default except at the start of the regex; you would need something like .* to make it allow additional characters between them.)
Since the ^ is not at the start of the expression, its meaning is 'not'. So in this case it means that there should not be a dash there.
I need a regular expression to list accepted Version Numbers. ie. Say I wanted to accept "V1.00" and "V1.02". I've tried this "(V1.00)|(V1.01)" which almost works but then if I input "V1.002" (Which is likely due to the weird version numbers I am working with) I still get a match. I need to match the exact strings.
Can anyone help?
The reason you're getting a match on "V1.002" is because it is seeing the substring "V1.00", which is part of your regex. You need to specify that there is nothing more to match. So, you could do this:
^(V1\.00|V1\.01)$
A more compact way of getting the same result would be:
^(V1\.0[01])$
Do this:
^(V1\.00|V1\.01)$
(. needs to be escaped, ^ means must be on the beginning of the text and $ must be on the end of the text)
I would use the '^' and '$' to mark the beginning and end of the string, like this:
^(V1\.00|V1\.01)$
That way the entire string must match the regex.