My problem is simple, but I've been pulling my hair out trying to solve it. I have two types of strings: one has a semicolon and the other doesn't. Both have colons.
Reason: A chosen reason
Delete: Other: testing
Reason for action: Other; testing
Blah: Other; testing;testing
If the string has a semicolon, I want to match anything after the first one. If it has no semicolon, I want to match everything after the first colon. For lines above I should get:
A chosen reason
Other: testing
testing
testing;testing
I can get the semicolon to match by using ;(.*) and I can get the colon to match by using :(.*).
I tried using an alternative like this: ;(.*)|:(.*) thinking that maybe if I have the right order I can get it to match the semicolon first, and then the colon if there is no semicolon, but it always just matched the colon.
What am I doing wrong?
Edit
I added another test case above to match the requirements I had stated. For strings with no semicolon, it should match the first colon.
Also, "Reason" could be anything, so I am clarifying that as well in the test cases.
Second Edit
To clarify, I'm using the POSIX Regular Expressions (using in PostgeSQL).
My guess is that you might want to design an expression, maybe similar to:
:\s*(?:[^;\r\n]*;)?\s*(.*)$
Demo
Here you have a fast regex (233 steps) with no look aheads.
.*?:\s*(?:([^\n;]+)|.*?;\s*(.*))$
Check out the regex https://regex101.com/r/9gbpjW/3
UPDATED: to match any placeholder. Instead of just Reason
One option is to use an alternation to first check if the string has no ; If there is none, then match until the first : and capture the rest in group 1.
In the case that there a ; match until the first semicolon and capture the rest in group 1.
For the logic stated in the question:
If the string has a semicolon, I want to match anything after the first one.
If it has no semicolon, I want to match everything after the first colon
You could use:
^(?:(?!.*;)[^\r\n:]*:|[^;\r\n]*;)[ \t]*(.*)$
Explanation
^ Start of string
(?: Non capturing group
(?!.*;) Negative lookahead (supported by Postgresql), assert string does not contain ;
[^\r\n:]*: If that is the case, match 0+ times not : or a newline, then match :
| Or
[^;\r\n]*; Match 0+ times not ; or newline, then match ;
) Close non capturing group
[ \t]* Match 0+ spaces or tabs
(.*) Capturing group 1, match any char 0+ times
$ End of string
Regex demo | Postgresql demo
regex = .*?:(?(?!.*;)(.*)|.*?;(.*))
demo
Related
I've a list below:
7080508136242611718:7080508978035787525:7549dda86ba9af19:31050:install_id=7080508978035787525; store-country-code=us; store-idc=useast5; ttreq=1$fd2f36282a10633c5638a02cc54c19ff13f60755; passport_csrf_token=13bf74c4e5fe04307f0a99de9aed53f9; passport_csrf_token_default=13bf74c4e5fe04307f0a99de9aed53f9; odin_tt=11ed1b48fba2d7a9fe3d86929b3d52cebbad0ca7f7dbd127e220cfb3be279621ba04487517b536050a6ded9fbe50e300cd11615e2e9551523478e5484896a9dda800e55e428842872fcf862e8c57d439:1648559503:351451268482810:3f:49:8c:b7:8c:cb:c5379d41-6cf3-4152-9d48-7aa45f7f611c:79375640-197c-4aaa-86cf-4ef8e7238be2:1:AgICAw0AFockF-RPsNA-7qeIMtk5-CKdkW2eP4TZYMDY7A
7080507996291827206:7080508977079666438:6742591cc0d20580:31050:install_id=7080508977079666438; store-country-code=us; store-idc=useast5; ttreq=1$a119611bfe79541b0b4c029fe910b6507123eec2; passport_csrf_token=fb42bbd472462c17f45acb531deb057a; passport_csrf_token_default=fb42bbd472462c17f45acb531deb057a; odin_tt=6c3b06ff01fd67f42e3dccb60a1e69ca67cb8654f49662017acc209f7176517bcd13a374311f7a1b3538e6407fb237267abf43578d3180d8c834e7df886fa4377a9b950dbb6ff146e3fabf37158dcfa8:1648559508:351451233766930:dd:9e:82:59:5f:7f:596da881-89e8-4f60-b644-5fef23f0a422:f04adc87-56de-4191-a25f-843bec1d5818:1:AgICAw0AFockF-RPsNA-7qeIMtk5-CKdsYPWv4TZYMDY7A
7080509102451394054:7080509820378072837:e36dc9aceecfc1cc:31050:install_id=7080509820378072837; store-country-code=us; store-idc=useast5; ttreq=1$d94700921d5ee2b21992910a2a4e84dd0ade1ec8; passport_csrf_token=2d4f4eca772dbfcbb37548ff02da3166; passport_csrf_token_default=2d4f4eca772dbfcbb37548ff02da3166; odin_tt=53d6999ebe29c0d5144a9669331ce3307a290891370914dabadbfa0520114e6e76b9103c9a6db5476e139251ee478f3a305577a89e3fa07288b7aca00774d3fccbd03566687dbcfdce31700065295939:1648559700:351451299637010:71:de:41:2b:ad:b4:1eba1ae9-3216-40e1-be7f-00303e524c27:2713cbd3-7a4f-493e-b76f-ac6d56ab8045:5:AgMNAgIAhyQWF-RPsNA-7qeIMtk5-CKcsBcWP4TZYMDY7w
7080509086894851590:7080509909225604870:98be64e38551984d:31050:install_id=7080509909225604870; store-country-code=us; store-idc=useast5; ttreq=1$05929375d8605739d8ebdbb5ce15eb406da5c467; passport_csrf_token=c95c71ad206a1d371e5b67505ae25be8; passport_csrf_token_default=c95c71ad206a1d371e5b67505ae25be8; odin_tt=6ddaa02f6133e61a4c591ef2a872f0ec2339d8b6a3fc480575fe279b13ded615e1fa7de979e18565f3ac8b8229a19a98bdf79aa1804071dcc025e1a4cd5314522cf40a62ca961770baea1d5d653d6d64:1648559720:351451292934660:9d:cf:c3:92:f6:f5:787dfb42-f4bf-43fa-9c64-ded19a1b1660:366c3024-217d-4f85-90dd-d95a0fd3e296:4:AgICAw0AFockF-RPsNA-7qeIMtk5-CKcs7bUP4TZYMDY7w
7080509183397299718:7080509974838085382:f39db5d314071713:31050:install_id=7080509974838085382; store-country-code=us; store-idc=useast5; ttreq=1$561ee2083cb13f0849a9f09e7f89edfe08c7ce6c; passport_csrf_token=721a8fee6f4f97c16ed1923ad3bbc72d; passport_csrf_token_default=721a8fee6f4f97c16ed1923ad3bbc72d;
I'd like to extract first two options aka below:
7080508136242611718:7080508978035787525
7080507996291827206:7080508977079666438
7080509102451394054:7080509820378072837
7080509086894851590:7080509909225604870
7080509183397299718:7080509974838085382
I've tried: *.: but its remove the reset of text. and keeps only first.
I've tried ^.*[0-9]+.*$ to get the second one. but no success.
Hopefully somebody can help me with accurate regex.
Thank you in advance.
This pattern *.: by itself is not a valid regex, and this pattern ^.*[0-9]+.*$ matches the whole string with at least a single digit.
If you want to match the digits and : you could make use of \K to forget what is matched so far and then match the rest of the line.
In the replacement use an empty string.
^\d+:\d+\K.*
^ Start of string
\d+:\d+ Match 1+ digits with : in between
\K.* Clear the current match, and match the rest of the line
Regex demo
^[^:]*:[^:]*\K.*
When matching things with delimiters I will use a negated character set to match the contents. In this case, the delimiter is a colon, so I want to match everything that isn't a colon until there's a colon. Then I want to match everything that isn't a colon. This will match everything up until the second colon. Because I want to keep what I just matched, I am using .* after \K, which resets the match at that point and matches everything else.
That pattern can be replaced with nothing, and the result is the first two columns of each line left.
You can use
Find: ^(\d+:\d+).*
Replace: $1
See this regex demo online.
The ^(\d+:\d+).* regex matches and captures into Group 1 one or more digits + : + one or more digits (with (\d+:\d+)) at the beginning of a line (^) and then matches the rest of the line (with .*).
The $1 replacement replaces the match with the Group 1 value.
See the demo and settings screenshot:
As an alternative, if there are chars other than digits you can also use
^([^:\v]+:[^:\v]+).*
where [^:\v]+ matches one or more chars other than a comma and any vertical whitespace.
I have a string and would like to match a part of it.
The string is Accept: multipart/mixedPrivacy: nonePAI: <sip:4168755400#1.1.1.238>From: <sip:4168755400#1.1.1.238>;tag=5430960946837208_c1b08.2.3.1602135087396.0_1237422_3895152To: <sip:4168755400#1.1.1.238>
I want to match PAI: <sip:4168755400#
the whitespace can be a word so i would like to use .* but if i used that it matches most of the string
The example on that link is showing what i'm matching if i use the whitespace instead of .*
(PAI: <sip:)((?:\([2-9]\d{2}\)\ ?|[2-9]\d{2}(?:\-?|\ ?))[2-9]\d{2}[- ]?\d{4})#
The example on that link is showing what i'm trying to achieve with .* but it should only match PAI: <sip:4168755400#
(PAI:.*<sip:)((?:\([2-9]\d{2}\)\ ?|[2-9]\d{2}(?:\-?|\ ?))[2-9]\d{2}[- ]?\d{4})#
I tried lookaround but failing.
Any idea?
thanks
Matching the single space can be updated by using a character class matching either a space or a word character and repeat that 1 or more times to match at least a single occurrence.
Note that you don't have to escape the spaces, and in both occasions you can use an optional character class matching either a space or hyphen [ -]?
If you want the match only, you can omit the 2 capturing groups if you want to.
(PAI:[ \w]+<sip:)((?:\([2-9]\d{2}\) ?|[2-9]\d{2}[ -]?)[2-9]\d{2}[- ]?\d{4})#
Regex demo
The regex should be like
PAI:.*?(<sip:.*?#)
Explanation:
PAI:.*? find the word PAI: and after the word it can be anything (.*) but ? is used to indicate that it should match as few as possible before it found the next expression.
(<sip:.*?#) capturing group that we want the result.
<sip:.*?# find <sip: and after the word it can be anything .*? before it found #.
Example
I have a comma separated string which I want to validate using a regex. What I have written is gives me a match if there a part wrong later in the string. I want to discard it completely if any part is wrong.
My regex : ^(?:[\w\.]+,{1}(?:STR|INT|REAL){1},{1}(\s*|$))+
Positive Case : Component,STR,YoungGenUse,STR,YoungGenMax,STR,OldGenUse,INT,OldGenMax,INT,PermGenUse,INT,PermGenMax,INT,MajCollCnt,INT,MinCollDur,REAL,MinCollCnt,INT,
Negative Case :
Component,STR,YoungGenUse,STR,YoungGenMax,TEST,OldGenUse,INT,OldGenMax,INT,PermGenUse,INT,PermGenMax,INT,MajCollCnt,INT,MinCollDur,REAL,MinCollCnt,INT,
For the second case, my regex gives a match for the bold portion eventhough, later there is an incorrect part (TEST). How can I modify my regex to discard the entire string?
The pattern that you tried would not match TEST in YoungGenMax,TEST because the alternatives STR|INT|REAL do not match it.
It would show until the last successful match in the repetition which would be Component,STR,YoungGenUse,STR,
You have to add the anchor at the end, outside of the repetition of the group, to indicate that the whole pattern should be followed by asserting the end of the string.
There are no spaces or dots in your string, so you might leave out \s* and use \w+ without the dot in the character class. Note that \s could also possibly match a newline.
^(?:\w+,(?:STR|INT|REAL),)+$
Regex demo
If you want to keep matching optional whitespace chars and the dot:
^(?:[\w.]+,(?:STR|INT|REAL),\s*)+$
Regex demo
Note that by repeating the group with the comma at the end, the string should always end with a comma. You can omit {1} from the pattern as it is superfluous.
your regex must keep matching until end of the string, so you must use $ to indicate end of the line:
^(?:[\w.]+,{1}(?:STR|INT|REAL){1},{1}(\s*|$))+$
Regex Demo
I am looking for regex to match following set:
/VIDEO_PRE_MINE
/VIDEO_PRE
/VIDEO_PRE/
/VIDEO_PRE/SOMETHING
And I want exclude expresions like this:
/VIDEO_PRESOMETHING
/VIDEO_PREsomething/something
In other words after expression '_PRE' cannot be any literal character, but it can be end of the string.
Here are regexes that i tried:
1. ^\/[^\/]*_PRE[^a-z|A-Z]
2. ^\/[^\/]*_PRE[^a-z|A-Z]?$
However I didn't manage to cover all use cases from sets with those regex.
I would really appreciate any help with this.
Thanks
For your example data, you could add an optional group (?:[_/].*)? to match either a _ or / followed by matching any char except a newline 0+ times until the end of the string $
^/[^/]*_PRE(?:[_/].*)?$
^ Start of string
/[^/]* Match /, then 0+ times any char except /
_PRE Match literally
(?: Non capturing group
[_/].* Match either _ or / followed by 0+ times any char except a newline
)? Close non capturing group and make it optional
$ End of string
Regex demo
Note that the forward slashes are not escaped. Depending on the language or delimiters you might have to escape them.
My guess is that we might want to have some right boundaries, such as
^\/VIDEO_PRE(?:\b\/?|\/[^\/\s]+\/?|_[^\/\s]+\/?)$
in specified form, and in general form:
^\/[^_]+_PRE(?:\b\/?|\/[^\/\s]+\/?|_[^\/\s]+\/?)$
which might work. You would likely want to test and modify the expression, which is explained on the top right panel of regex101.com, if you wish to explore/simplify it, and in this link, you can watch how it would match against some sample inputs, if you like.
DEMO
I need to create regex to find last underscore in string like 012344_2.0224.71_3 or 012354_5.00123.AR_3.335_8
I have wanted find last part with expression [^.]+$ and then find underscore at found element but I can not handle it.
I hope you can help me :)
Just use a negative character class [^_] that will match everything except an underscore (this helps to ensure no other underscores are found afterwards) and end of string $
Pattern would look as such:
(_)[^_]*$
The final underscore _ is in a capturing group, so you are wanting to return the submatch. You would replace the group 1 (your underscore).
See it live: Regex101
Notice the green highlighted portion on Regex101, this is your submatch and is what would be replaced.
The simplest solution I can imagine is using .*\K_, however not all regex flavours support \K.
If not, another idea would be to use _(?=[^_]*$)
You have a demo of the first and second option.
Explanation:
.*\K_: Fetches any character until an underscore. Since the * quantifier is greedy, It will match until the last underscore. Then \K discards the previous match and then we match the underscore.
_(?=[^_]*$): Fetch an underscore preceeded by non-underscore characters until the end of the line
If you want nothing but the "net" (i.e., nothing matched except the last underscore), use positive lookahead to check that no more underscores are in the string:
/_(?=[^_]*$)/gm
Demo
The pattern [^.]+$ matches not a dot 1+ times and then asserts the end of the string. The will give you the matches 71_3 and 335_8
What you want to match is an underscore when there are no more underscores following.
One way to do that is using a negative lookahead (?!.*_) if that is supported which asserts what is at the right does not match any character followed by an underscore
_(?!.*_)
Pattern demo