So I am trying to make a regex match for strings of the form:
"catalog.schema.'tablename'" .
The output I am looking for is just catalog.schema.'tablename' leaving out the quotes at the end position.
Can anyone help me out
I tried to do it with the expression
/(?!^|.$)+[^\s]/ which leaves out the end quotes but matches each character.
So I modified it to /(?!^|.$)+[^\s]+/g . This matches the whole sentence but doesn't ignore the end quote.
Depends on the data arround your string and quotationmarks may be within the string.
Why not just this: "(.*?)"
https://regex101.com/r/oaS8o0/1
To answer the question in the title you might simply use:
^.(.*)?.$
https://regex101.com/r/FxJgtW/1
You can just use
(?<=.).+(?=.)
Or, if you cannot use lookbehind:
(?!^).+(?!$)
See the regex demo #1 and regex demo #2.
Since . matches any char other than line break chars, the patterns just match any strings without their start and end chars.
If you don't want to match the first and the last character, you can just use a capture group instead of lookarounds and use the group 1 value.
The first . matches the first of (any) characters, the (.+) is a capture group that matches 1 or more characters, and the . at the end matches the last character of the string.
.(.+).
Regex demo
Or to get the text between the double quotes at the start and the end of the string using a negated character class and a capture group:
^"([^"]+)"$
Regex demo
Related
I just started learning regex and I'm trying to understand how it possible to do the following:
If I have:
helmut_rankl:20Suzuki12
helmut1195:wasserfall1974
helmut1951:roller11
Get:
helmut_rankl:20Suzuki1
helmut1195:wasserfall197
helmut1951:roller1
I tried using .$ which actually match the last character of a string, but it doesn't match letters and numbers.
How do I get these results from the input?
You could match the whole line, and assert a single char to the right if you want to match at least a single character.
.+(?=.)
Regex demo
If you also want to match empty strings:
.*(?=.)
This will do what you want with regex's match function.
^(.*).$
Broken down:
^ matches the start of the string
( and ) denote a capturing group. The matches which fall within it are returned.
.* matches everything, as much as it can.
The final . matches any single character (i.e. the last character of the line)
$ matches the end of the line/input
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 $1 replacement replaces the match with the Group 1 value.
See the demo and settings screenshot:
As an alternative, if there are chars other than digits you can also use
^([^:\v]+:[^:\v]+).*
where [^:\v]+ matches one or more chars other than a comma and any vertical whitespace.
I thought I understood how the non-greedy modifier works, but am confused by the following result:
Regular Expression: (,\S+?)_sys$
Test String: abc,def,ghi,jkl_sys
Desired result: ,jkl_sys <- last field including comma
Actual result: ,def,ghi,jkl_sys
Use case is that I have a comma separated string whose last field will end in "_sys" (e.g. ,sometext_sys). I want to match only the last field and only if it ends with _sys.
I am using the non-greedy (?) modifier to return the shortest possible match (only the last field including the comma), but it returns all but the first field (i.e. the longest match).
What am I missing?
I used https://regex101.com/ to test, in case you want to see a live example.
You can use
,[^,]+_sys$
The pattern matches:
, Match the last comma
[^,]+ Match 1 + occurrences of any char except ,
_sys Match literally
$ End of string
See a regex demo.
If you don't want to match newlines and whitespaces:
,[^\s,]+_sys$
It sounds like you're looking for the a string that ends with "_sys" and it has to be at the end of the source string, and it has to be preceded by a comma.
,\s*(\w+_sys)$
I added the \s* to allow for optional whitespace after the comma.
No non-greedy modifiers necessary.
The parens are around \w+_sys so you can capture just that string, without the comma and optional whitespace.
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
Take this string
bob.ted.dave.allan::james.fred: hello.dave
I need to replace each occurrence of . with # until I hit a singular ":" not stopping at the namespace '::' but just the ":"
So in the above string, the end result will be
bob#ted#dave#allan::james#fred: hello.dave
currently I have [.] which matches all dots in the string but cant get it to anchor on the single ":"
If you want a pure regex solution this can only be done with a variable-width lookbehind pattern, which is supported by only a few regex engines:
(?<!(?<!:):(?!:).*)\.
Demo: https://regex101.com/r/Crq49C/2
Or if there is always going to be a colon, you can use a positive lookahead pattern instead:
\.(?=.*(?<!:):(?!:))
Demo: https://regex101.com/r/Crq49C/3
To just anchor on a single . and not stop until seeing exactly one : I think
[.](?=.*:{1})
will work.
This does not work if there is more than one : on a line.
You can capture the entire string up until the single : with a simple character set match:
[\w\. :]+(?::{1})
Then do a replace on all of the .'s in the captured string.
Demo: https://regex101.com/r/xDfstu/1