Develop Reference

Regex - extract string between double quotes where line starts with a keyword

Given the following lines:
classes!("text-xl", "text-blue-500")
String::from("test")
classes!("text-sm")
I wish to extract text-xl, text-blue-500 and text-sm and potentially any other strings within classes!
What I have so far is:
(?<=classes!).* - this gives me everything after classes! but not
the exact values
(?!^")".*?" - this gives me the values I want with but with the quotes
I use the global and multi line flags

If you are using Javascript, and the positive lookbehind is supported (you don't need the multiline flag, only the global flag)
(?<=classes!\((?:"[^"]*",\s*)*")[^"]*(?=")
(?<= Positive lookbehind to assert to the left
classes!\( Match classes!(
(?:"[^"]*",\s*)* Match optional repetitions of "...",
" Match a double quote
) Close lookbehind
[^"]* Match optional chars other than "
(?=") Assert " to the right
Regex demo
const regex = /(?<=classes!\((?:"[^"]*",\s*)*")[^"]*(?=")/g;
const str = `classes!("text-xl", "text-blue-500")
String::from("test")
classes!("text-sm")
`;
console.log(str.match(regex))

You could try a non fixed-width lookbehind like so:
(?<=^classes!\("(?:[^"]*"[^"()]*")*)[^"]*
See an online demo
(?<= - Open a positive lookbehind;
^classes!\(" - Start-line anchor and match literally 'classes!("';
(?:[^"]*"[^()"]*")* - A nested non-capture group to match (0+ times) any char other than quotes (and paranthesis), a quote and that same sequence repeated.
) - Close lookbehind;
[^"]* - Match 0+ (Greedy) characters other than quotes.

Regex to match string inside square brackets, separated by colon

I'm trying to parse a string that always has the format: [firstpart:lastpart] in such a way that I can get "firstpart" and "lastpart" as separate items. The "firstpart" value is always a string, and the "lastpart" value could contain integers and text. The whole string [firstpart:lastpart] could be surrounded by any amount of other text that I don't need, hence the brackets.
I've been trying to modify this:
([^:\s]+):([^:\s]+)
As is, it gets me this:
[firstpart:lastpart
[firstpart
lastpart]
So it's just that I need to remove the open and close brackets from 2 and 3.
Is this possible with just a regex? I'm using JavaScript in a TinyMCE plugin, in case that is relevant.

Put \[ and \] at the beginning and end of the regular expression, respectively, and capture the text between them:
console.log(
'foo[firstpart:lastpart]bar'.match(/\[([^:]+):([^:\]]+)\]/)
);

You could match the opening and the closing bracket outside of the group:
\[([a-z]+):([a-z0-9]+)]
Note that [^:\s]+ Matches not a colon or a whitespace character which matches more than a string or a string or integers and escape the opening \[ to match it literally or else it would start a character class.
let str = "[firstpart:lastpart]";
console.log(str.match(/\[([a-z]+):([a-z0-9]+)]/i));

How to replace different characters with regex and add conditionals.

Example string: George's - super duper (Computer)
Wanted new string: georges-super-duper-computer
Current regex: .replace(/\s+|'|()/g, '-')
It does not work and and when I remove the spaces and there is already a - in between I get something like george's---super.

tl;dr Your regex is malformed. Also you can't conditionally remove ' and \s ( ) in a single expression.
Your regex is malformed since ( and ) have special meanings. They are used to form groups so you have to escape them as \( and \). You'll also have to place another pipe | in between them, otherwise you're going to match the literal "()", which is not what you want.
The proper expression would look like this: .replace(/\s+|'|\(|\)/g, '-').
However, this is not what you want. Since this would produce George-s---super-duper--Computer-. I would recommend that you use Character Classes, which will also make your expression easier to read:
.replace(/[\s'()-]+/g, '-')
This matches whitespace, ', (, ) and any additional - on or more times and replaces them with -, yielding George-s-super-duper-Computer-.
This is still not quite right, so have this:
var myString = "George's - super duper (Computer)";
var myOtherString = myString
// Remove non-whitespace, non-alphanumeric characters from the string (note: ^ inverses the character class)
// also trim any whitespace from the beginning and end of the string (so we don't end up with hyphens at the start and end of the string)
.replace(/^\s+|[^\s\w]+|\s+$/g, "")
// Replace the remaining whitespace with hyphens
.replace(/\s+/g, "-")
// Finally make all characters lower case
.toLowerCase();
console.log(myString, '=>', myOtherString);

You could do match instead of replace then join result on -. Then you may need a replace to remove single quotes. Regex would be:
[a-z]+('[a-z]+)*
JS code:
var str = "George's - super duper (Computer)";
console.log(
str.match(/[a-z]+('[a-z]+)*/gi).join('-').replace("'", "").toLowerCase()
);

Split string on spaces except for in quotes, but include incomplete quotes

I am trying to split a string in JS on spaces except when the space is in a quote. However, an incomplete quote should be maintained. I'm not skilled in regex wizardry, and have been using the below regex:
var list = text.match(/[^\s"]+|"([^"]*)"/g)
However, if I provide input like sdfj "sdfjjk this will become ["sdfj","sdfjjk"] rather than ["sdfj",""sdfjjk"].

You can use
var re = /"([^"]*)"|\S+/g;
By using \S (=[^\s]) we just drop the " from the negated character class.
By placing the "([^"]*)" pattern before \S+, we make sure substrings in quotes are not torn if they come before. This should work if the string contains well-paired quoted substrings and the last is unpaired.
Demo:
var re = /"([^"]*)"|\S+/g;
var str = 'sdfj "sdfjjk';
document.body.innerHTML = JSON.stringify(str.match(re));
Note that to get the captured texts in-between quotes, you will need to use RegExp#exec in a loop (as String#match "drops" submatches).
UPDATE
No idea what downvoter thought when downvoting, but let me guess. The quotes are usually used around word characters. If there is a "wild" quote, it is still a quote right before/after a word.
So, we can utilize word boundaries like this:
"\b[^"]*\b"|\S+
See regex demo.
Here, "\b[^"]*\b" matches a " that is followed by a word character, then matches zero or more characters other than " and then is followed with a " that is preceded with a word character.
Moving further in this direction, we can make it as far as:
\B"\b[^"\n]*\b"\B|\S+
With \B" we require that " should be preceded with a non-word character, and "\B should be followed with a non-word character.
See another regex demo
A lot depends on what specific issue you have with your specific input!

Try the following:
text.match(/".*?"|[^\s]+/g).map(s => s.replace(/^"(.*)"$/, "$1"))
This repeatedly finds either properly quoted substrings (first), OR other sequences of non-whitespace. The map part is to remove the quotes around the quoted substrings.
> text = 'abc "def ghi" lmn "opq'
< ["abc", "def ghi", "lmn", ""opq"]

How can I put [] (square brackets) in RegExp javascript?

I'm trying this:
str = "bla [bla]";
str = str.replace(/\\[\\]/g,"");
console.log(str);
And the replace doesn't work, what am I doing wrong?
UPDATE: I'm trying to remove any square brackets in the string,
what's weird is that if I do
replace(/\[/g, '')
replace(/\]/g, '')
it works, but
replace(/\[\]/g, ''); doesn't.

It should be:
str = str.replace(/\[.*?\]/g,"");
You don't need double backslashes (\) because it's not a string but a regex statement, if you build the regex from a string you do need the double backslashes ;).
It was also literally interpreting the 1 (which wasn't matching). Using .* says any value between the square brackets.
The new RegExp string build version would be:
str=str.replace(new RegExp("\\[.*?\\]","g"),"");
UPDATE: To remove square brackets only:
str = str.replace(/\[(.*?)\]/g,"$1");
Your above code isn't working, because it's trying to match "[]" (sequentially without anything allowed between). We can get around this by non-greedy group-matching ((.*?)) what's between the square brackets, and using a backreference ($1) for the replacement.
UPDATE 2: To remove multiple square brackets
str = str.replace(/\[+(.*?)\]+/g,"$1");
// bla [bla] [[blaa]] -> bla bla blaa
// bla [bla] [[[blaa] -> bla bla blaa
Note this doesn't match open/close quantities, simply removes all sequential opens and closes. Also if the sequential brackets have separators (spaces etc) it won't match.

You have to escape the bracket, like \[ and \]. Check out http://regexpal.com/. It's pretty useful :)
To replace all brackets in a string, this should do the job:
str.replace(/\[|\]/g,'');
I hope this helps.
Hristo

Here's a trivial example but worked for me. You have to escape each sq bracket, then enclose those brackets within a bracket expression to capture all instances.
const stringWithBrackets = '[]]][[]]testing][[]][';
const stringWithReplacedBrackets = stringWithBrackets.replace(/[\[\]]/g, '');
console.log(stringWithReplacedBrackets);

Two backslashes produces a single backslash, so you're searching for "a backslash, followed by a character class consisting of a 1 or a right bracket, and then you're missing an closing bracket.
Try
str.replace(/\[1\]/g, '');

What exactly are you trying to match?
If you don't escape the brackets, they are considered character classes. This:
/[1\\]/
Matches either a 1 or a backslash. You may want to escape them with one backslash only:
/\[1\]/
But this won't match either, as you don't have a [1] in your string.

I stumbled on this question while dealing with square bracket escaping within a character class that was designed for use with password validation requiring the presence of special characters.
Note the double escaping:
var regex = new RegExp('[\\]]');
As #rudu mentions, this expression is within a string so it must be double escaped. Note that the quoting type (single/double) is not relevant here.
Here is an example of using square brackets in a character class that tests for all the special characters found on my keyboard:
var regex = new RegExp('[-,_,\',",;,:,!,#,#,$,%,^,&,*,(,),[,\\],\?,{,},|,+,=,<,>,~,`,\\\\,\,,\/,.]', 'g')

How about the following?
str = "bla [bla]";
str.replace(/[[\\]]/g,'');
You create a character set with just the two characters you are interested in and do a global replace.

Nobody quite made it simple and correct:
str.replace(/[[\]]/g, '');
Note the use of a character class, with no escape for the open bracket, and a single backslash escape for the close bracket.