I'm trying to create a regex for extracting singers, lyricists. I was wondering how to make lyricists search optional.
Sample Multiline String:
Fireworks Singer: Katy Perry
Vogue Singers: Madonna, Karen Lyricist: Madonna
Regex: /Singers?:(.\*)\s?Lyricists?:(.\*)/
This matches the second line correctly and extracts Singers(Madonna, Karen) and Lyricists(Madonna)
But it does not work with the first line, when there are no Lyricists.
How do I make Lyricists search optional?
You can enclose the part you want to match in a non-capturing group: (?:). Then it can be treated as a single unit in the regex, and subsequently you can put a ? after it to make it optional. Example:
/Singers?:(.*)\s?(?:Lyricists?:(.*))?/
Note that here the \s? is useless since .* will greedily eat all characters, and no backtracking will be necessary. This also means that the (?:Lyricists?:(.*)) part will never be matched for the same reason. You can use the non-greedy version of .*, .*? along with the $ to fix this:
/Singers?:(.*?)\s*(?:Lyricists?:(.*))?$/
Some extra whitespace ends up captured; this can be removed also, giving a final regex of:
/Singers?:\s*(.*?)\s*(?:Lyricists?:\s*(.*))?$/
Just to add to Cameron's solution. if the source string has multiple lines each containing both Singers and Lyricists, you'll probably need to add the 'm' multi-line modifier so that the '$' will match ends-of-lines. (You didn't say what language you are using - you may want to add the 'i' modifier as well.)
Related
I've been trying hard to get this Regex to work, but am simply not good enough at this stuff apparently :(
Regex - Trying to extract sources
I thought this would work... I'm trying to get all of the content where:
It starts with ds://
Ends with either carriage return or line feed
That's it! Essentially I'm going to then do a negative lookahead such that I can remove all content that is NOT conforming to above (in Notepad++) which allows for Regex search/replace.
Search for lines that contain the pattern, and mark them
Search menu > Mark
Find what: ds://.*\R
check Regular expression
Check Mark the lines
Find all
Remove the non marked lines
Search menu > Bookmark
Remove unmarked lines
You don't need to add the \w specifier to look for a word after the ds:// in the look ahead. Removing that and altering the final specification from "zero or one carriage return, then zero or one newline" to "either a carriage return or a newline" in capture group should do it for you:
(?=ds:\/\/).*(?:\r|\n)
Update: Carriage return or Line feed group does not need to be captured.
Update 2: The following regex will actually work for your proposed use case in the comments, matching everything but the pattern you described in the question.
^(?:(?!ds:\/\/.*(?:\r|\n)).)*$
You regex (?=ds:\w+).*\r?\n? does not match because in the content there is ds:// and \w does not match a forward slash. To make your regex work you could change it to:
(?=ds://\w+).*\r?\n? demo which can be shortened to ds://.*\R? demo
Note that you don't have to escape the forward slash.
If you want to do a find and replace to keep the lines that contain ds:// you could use a negative lookahead:
Find what
^(?!.*ds://).*\R?
Replace with
Leave empty
Explanation
^ Start of the string
(?!.*ds://) Negative lookahead to assert the string does not contain ds://
.* Match any character 0+ times
\R? An optional unicode newline sequence to also match the last line if it is not followed by a newline
See the Regex demo
Here you go, Andrew:
Regex: ds:\/\/.*
Link: https://regex101.com/r/ulO9GO/2
Let me know if any question.
In the line below, I need to these two lines into one single line by replacing the newline and empty space with nothing.
Provisioned Links : 2/14, 2/24, 7/10, 7/12,
7/25, 7/31, 7/32
Therefore I have this regex (in Notepad++):
(\r\n|\n)\s+[0-9]\/[0-9]*
Problem: the match includes the 7/25 - I need it to look for the #/## but not include it.
If I use this lookaround pattern:
(\r\n|\n)\s+(q=[0-9]\/[0-9])*
all lines beginning with newline + spaces are matched, whether or not they end with #/##.
What am I doing wrong?
regex101 fiddle to play with
Be careful:
You should correct the way you constructed the lookahead: (?=....)
Lookarounds are not quantifiable.
so what you need really is [\r\n]\s+(?=[0-9]\/[0-9]*).
Live demo
To normalize whitespace, why not simply replace "comma with additional space after it" with "comma plus one tab character" ?
You don't need that complicated pattern at all, because \s matches spaces, newlines, and tabs all at the same time:
Pattern: ,\s*
Replacement string: ,\t
https://regex101.com/r/T0QJnq/1
I'm using Regex to match whole sentences in a text containing a certain string. This is working fine as long as the sentence ends with any kind of punctuation. It does not work however when the sentence is at the end of the text without any punctuation.
This is my current expression:
[^.?!]*(?<=[.?\s!])string(?=[\s.?!])[^.?!]*[.?!]
Works for:
This is a sentence with string. More text.
Does not work for:
More text. This is a sentence with string
Is there any way to make this word as intended? I can't find any character class for "end of text".
End of text is matched by the anchor $, not a character class.
You have two separate issues you need to address: (1) the sentence ending directly after string, and (2) the sentence ending sometime after string but with no end-of-sentence punctuation.
To do this, you need to make the match after string optional, but anchor that match to the end of the string. This also means that, after you recognize an (optional) end-of-sentence punctuation mark, you need to match everything that follows, so the end-of-string anchor will match.
My changes: Take everything after string in your original regex and surround it in (?:...)? - the (?:...) being a "non-remembered" group, and the ? making the entire group optional. Follow that with $ to anchor the end of the string.
Within that optional group, you also need to make the end-of-sentence itself optional, by replacing the simple [.?!] with (?:[.?!].*)? - again, the (?:...) is to make a "non-remembered" group, the ? makes the group optional - and the .* allows this to match as much as you want after the end-of-sentence has been found.
[^.?!]*(?<=[.?\s!])string(?:(?=[\s.?!])[^.?!]*(?:[.?!].*)?)?$
The symbol for end-of-text is $ (and, the symbol for beginning-of-text, if you ever need it, is ^).
You probably won't get what you're looking for with by just adding the $ to your punctuation list though (e.g., [.?!$]); you'll find it works better as an alternative choice: ([.?!]|$).
Your regex is way too complex for what you want to achieve.
To match only a word just use
"\bstring\b"
It will match start, end and any non-alphanum delimiters.
It works with the following:
string is at the start
this is the end string
this is a string.
stringing won't match (you don't want a match here)
You should add the language in the question for more information about using.
Here is my example using javascript:
var reg = /^([\w\s\.]*)string([\w\s\.]*)$/;
console.log(reg.test('This is a sentence with string. More text.'));
console.log(reg.test('More text. This is a sentence with string'));
console.log(reg.test('string'))
Note:
* : Match zero or more times.
? : Match zero or one time.
+ : Match one or more times.
You can change * with ? or + if you want more definition.
I'm trying to add another feature to a regex which is trying to validate names (first or last).
At the moment it looks like this:
/^(?!^mr$|^mrs$|^ms$|^miss$|^dr$|^mr-mrs$)([a-z][a-z'-]{1,})$/i
https://regex101.com/r/pQ1tP2/1
The idea is to do the following
Don't allow just adding a title like Mr, Mrs etc
Ensure the first character is a letter
Ensure subsequent characters are either letters, hyphens or apostrophes
Minimum of two characters
I have managed to get this far (shockingly I find regex so confusing lol).
It matches things like O'Brian or Anne-Marie etc and is doing a pretty good job.
My next additions I've struggled with though! trying to add additional features to the regex to not match on the following:
Just entering the same characters i.e. aaa bbbbb etc
Thanks :)
I'd add another negative lookahead alternative matching against ^(.)\1*$, that is, any character, repetead until the end of the string.
Included as is in your regex, it would make that :
/^(?!^mr$|^mrs$|^ms$|^miss$|^dr$|^mr-mrs$|^(.)\1*$)([a-z][a-z'-]{1,})$/i
However, I would probably simplify your negative lookahead as follows :
/^(?!(mr|ms|miss|dr|mr-mrs|(.)\2*)$)([a-z][a-z'-]{1,})$/i
The modifications are as follow :
We're evaluating the lookahead at the start of the string, as indicated by the ^ preceding it : no need to repeat that we match the start of the string in its clauses
Each alternative match the end of the string. We can put the alternatives in a group, which will be followed by the end-of-string anchor
We have created a new group, which we have to take into account in our back-reference : to reference the same group, it now must address \2 rather than \1. An alternative in certain regex flavours would have been to use a non-capturing group (?:...)
I'm trying to match first occurrence of window.location.replace("http://stackoverflow.com") in some HTML string.
Especially I want to capture the URL of the first window.location.replace entry in whole HTML string.
So for capturing URL I formulated this 2 rules:
it should be after this string: window.location.redirect("
it should be before this string ")
To achieve it I think I need to use lookbehind (for 1st rule) and lookahead (for 2nd rule).
I end up with this Regex:
.+(?<=window\.location\.redirect\(\"?=\"\))
It doesn't work. I'm not even sure that it legal to mix both rules like I did.
Can you please help me with translating my rules to Regex? Other ways of doing this (without lookahead(behind)) also appreciated.
The pattern you wrote is really not the one you need as it matches something very different from what you expect: text window.location.redirect("=") in text window.location.redirect("=") something. And it will only work in PCRE/Python if you remove the ? from before \" (as lookbehinds should be fixed-width in PCRE). It will work with ? in .NET regex.
If it is JS, you just cannot use a lookbehind as its regex engine does not support them.
Instead, use a capturing group around the unknown part you want to get:
/window\.location\.redirect\("([^"]*)"\)/
or
/window\.location\.redirect\("(.*?)"\)/
See the regex demo
No /g modifier will allow matching just one, first occurrence. Access the value you need inside Group 1.
The ([^"]*) captures 0+ characters other than a double quote (URLs you need should not have it). If these URLs you have contain a ", you should use the second approach as (.*?) will match any 0+ characters other than a newline up to the first ").