Develop Reference

Unable to match regex for any character except ' and "

I've written a regex to match against the string
{{AB.group.one}}:"eighth",{{AB.group.TWO}}:"third",{{attr1111}}:"fourth","fifth":{{attr_22_2qq2}},"sixth":{{AB.group.three}},{{ab.group.fourth}}:"seventh","ninth":{{attr1111}}}
Regex:
/[^'"]({{2}[a-zA-Z0-9$_].*?}{2})[^'"]/gi
Breaking the regex above:
[^'"]: Start with a character which is neither ' nor ".
({{2}[a-zA-Z0-9$_].*?}{2}): Have exactly 2 {{, then any character in the range a-zA-Z0-9$_ . After that, exactly 2 }}
[^'"]: Any character except for ' and ".
Below matches are not the exact matches but the captured groups. I'll perform my operations on the captured groups so for simplicity, we can consider them as our matches.
Expected matches:
{{AB.group.one}}
{{AB.group.TWO}}
{{attr1111}}
{{attr_22_2qq2}}
{{AB.group.three}}
{{ab.group.fourth}}
{{attr1111}}}
Resultant matches:
{{AB.group.TWO}}
{{attr1111}}
{{attr_22_2qq2}}
{{AB.group.three}}
{{attr1111}}}
As you can see in the image below {{AB.group.one}} and {{ab.group.fourth}} do not match. I want them to match them as well.
I know the reasons why they aren't matching.
The reason why {{AB.group.one}} doesn't match is because [^'"] expects one character except for ' and " and I'm not providing one. If I replace [^'"] with ["'"]*, it'll work but in that case "{{AB.group.one}}" will match as well.
So, the problem statement is match any character(if there's any) before {{ and after }} but the character can't be ' or ".
The reason why {{ab.group.fourth}} doesn't match is because the character preceding this match i.e. , is part of another match. This is just my speculation, the reason could be something else. But if I include any character between {{AB.group.three}}, and {{ab.group.fourth}} (e.g. {{AB.group.three}}, {{ab.group.fourth}}), then the pattern matches. I have no idea how can I fix this.
Please help me in solving these two problems. Thank you.

Here is a regex based approach which seems to be working. First, we can string off all double-quoted terms, then replace islands of comma/colon with just a single comma separator. Finally, split on comma to generate an array of terms.
var input = "{{AB.group.one}}:\"eighth\",{{AB.group.TWO}}:\"third\",{{attr1111}}:\"fourth\",\"fifth\":{{attr_22_2qq2}},\"sixth\":{{AB.group.three}},{{ab.group.fourth}}:\"seventh\",\"ninth\":{{attr1111}}},\"blah\":\"stuff\",{{one}}:{{two}}";
var terms = input.replace(/\".*?\"/g, "").replace(/[,:]+/g, ",").split(",");
console.log(terms);

You were actually really close with what you had.
let input = '{{AB.group.one}}:"eighth",{{AB.group.TWO}}:"third",{{attr1111}}:"fourth","fifth":{{attr_22_2qq2}},"sixth":{{AB.group.three}},{{ab.group.fourth}}:"seventh","ninth":{{attr1111}}}'
let regex = /(?<=[^'"]?)({{2}[a-zA-Z0-9$_].*?}{2})(?=[^'"]?)/gi;
console.log(input.match(regex))
(?<=[^'"]?) is a positive lookbehind. Since the negated character set is used, we're checking that the character before the match is not ' or ". The question mark makes this optional - match zero or one of the previous token (the negated character set).
(?=[^'"]?) is a positive lookahead and checks the token immediately after the expression to ensure that it's not a ' or " (or that there is no token after the expression).
Another option, since lookbehinds aren't supported in every browser:
let input = '{{AB.group.one}}:"eighth",{{AB.group.TWO}}:"third",{{attr1111}}:"fourth","fifth":{{attr_22_2qq2}},"sixth":{{AB.group.three}},{{ab.group.fourth}}:"seventh","ninth":{{attr1111}}}'
let regex = /(?:[^{'"])?({{2}[a-zA-Z0-9$_].*?}{2})(?:[^}'"])?/gi
console.log([...input.matchAll(regex)].map(reg => reg[1]))
String.match() loses reference to capture groups when the global flag is passed, so only returns the "match". Since you're creating a capture group with ({{2}[a-zA-Z0-9$_].*?}{2}), if you wanted to just ensure the characters immediately surrounding the bracketed expression aren't quotation marks, you can just use non-capture groups for those optional checks.
(?:[^{'"])? is a non-capturing group, as is (?:[^}'"])?
Using String.matchAll, the first element of the arrays created for each match is the entire match, the second element is the first capturing group, etc. So the logic for mapping over [...input.matchAll(regex)] is just to collect the capturing group from each match.

Javascript - Regular Expression with string template followed by 4 digits number?

Good day. I wanna detect the url string in the <a> tag
Link
whether it matchs the pattern : ?post_type=tribe_events&p=#### (#### = 4 digits number)
I'm writing some Jquery code to detect the expression but the console is throwing the error :
Invalid regular expression: /^(?)post_type=tribe_events&p=^(d{4})/:
Invalid group
var str = $(a).attr("href");
var regexEx = /^(?)post_type=tribe_events&p=^(d{4})/;
var ok = regexEx.exec(str);
console.log(ok);
I'm not good at the regex so I'd be aprreciated if there's any help.

There are couple of issues in your regex.
You need to remove ^ from your regex which denotes start of string and in your case your string doesn't actually start from a ? and is in middle of the string.
You need to escape ? as it has special meaning in regex which is zero or one occurrence of a character.
You need to remove second ^ after p= which isn't needed
You need to write \d and not just d for representing a number.
Also you don't need to group ? and \d{4} unless you really need them.
You corrected regex becomes,
\?post_type=tribe_events&p=\d{4}
Demo

If the test is really what you want, I suppose the right syntax would be:
/^\?post_type=tribe_events&p=\d{4}/

Regex : Prevent a string to end with specif special characters

I'm using reactive form validation under my Angular app .
i Have a form validation condition which demands that my string didn't end with
- or '
i ve tried this , but it seems to not work :
Validators.pattern('.*(?<!\-\')$')
Suggestions ?

The (?<!\-\') pattern is a negative lookbehind that matches a location that is not immediately preceded with a -' substring, while you only want to fail the match if there is either - or a ' at the end.
You may use the following solution that uses a regex literal notation, so that you didn't have to write a pattern to match the whole input:
Validators.pattern(/[^-']$/)
Or, if the string can be empty,
Validators.pattern(/(?:^|[^-'])$/)
The (?:^|[^-'])$ pattern matches either start of a string (^) or any char but a - or ' (with the help of the negated character class, [^-']), and $ asserts the end of the string.

How to replace a substring with open parentheses (

I am a Regex newbie and trying to implement Regex to replace a matching pattern in a string only when it has a ( - open parentheses using Javascript. for example if I have a string
IN(INTERM_LEVEL_IN + (int)X_ID)
I would only like to highlight the first IN( in the string. Not the INTERM_LEVEL_IN (2 ins here) and the int.
What is the Regex to accomplish this?

To match the opening bracket you just need to escape it: IN\(.
For instance, running this in Firebug console:
enter code here"IN(INTERM_LEVEL_IN + (int)X_ID)".replace(/(IN()/, 'test');`
Will result in:
>>> "IN(INTERM_LEVEL_IN + (int)X_ID)".replace(/(IN\()/, 'test');
"testINTERM_LEVEL_IN + (int)X_ID)"

Parenthesis in regular expressions have a special meaning (sub-capture groups), so when you want them to be interpreted literally you have to escape them by with a \ before them. The regular expression IN\( would match the string IN(.

The following should only match IN( at the beginning of a line:
/^IN\(/
The following would match IN( that is not preceded by any alphanumeric character or underscore:
/[a-zA-Z0-9_]IN\(/
And finally, the following would match any instance of IN( no matter what precedes it:
/IN\(/
So, take your pick. If you're interested in learning more about regex, here's a good tutorial: http://www.regular-expressions.info/tutorial.html

You can use just regular old Javascript for regex, a simple IN\( would work for the example you gave (see here), but I suspect your situation is more complicated than that. In which case, you need to define exactly what you are trying to match and what you don't want to match.

Javascript lookahead regular expression

I'm trying to write a regular expression to parse the following string out into three distinct parts. This is for a highlighting engine I'm writing:
"\nOn and available after solution."
I have a regular expression that's dynamically created for any word a user might input. In the above example, the word is "on".
The regular expression expects a word with any amount of white space ([\s]*) followed by the search word (with no -\w following it, eg: on-time, on-wards should not be a valid result. To complicate this, there can be a -,$,< or > symbol following the example, so on-, on> or on$ are valid. This is why there is a negative lookahead after the search word in my regular expression below.
There's a complicated reason for this, but it's not relevant to the question. The last part should be the rest of the sentence. In this example, " and available after solution."
So,
p1 = "\n"
p2 = "On"
p3 = " and available after solution"
I currently have the following regular expression.
test = new RegExp('([\\s]*)(on(?!\\-\\w))([$\\-><]*?\\s(?=[.]*))',"gi")
The first part of this regular expression ([\\s]*)(on(?!\\-\\w))[$\\-><]*? works as expected. The last part does not.
In the last part, what I'm trying to do is force the regular expression engine to match whitespace before matching additional characters. If it can not match a space, then the regular expression should end. However, when I run this regular expression, I get the following results
str1 = "\nOn ly available after solution."
test.exec(str1)
["\n On ", "\n ", "On"]
So it would appear to me that the last positive look ahead is not working. Thanks for any suggestions, and if anyone needs some clarification, let me know.
EDIT:
It would appear that my regular expression was not matching because I didn't realize the following caveat:
You can use any regular expression inside the lookahead. (Note that this is not the case with lookbehind. I will explain why below.) Any valid regular expression can be used inside the lookahead. If it contains capturing parentheses, the backreferences will be saved. Note that the lookahead itself does not create a backreference. So it is not included in the count towards numbering the backreferences. If you want to store the match of the regex inside a backreference, you have to put capturing parentheses around the regex inside the lookahead, like this: (?=(regex)). The other way around will not work, because the lookahead will already have discarded the regex match by the time the backreference is to be saved.

The dot in the character class [.] means a literal dot. Change it to just . if you wish to match any character.
The lookahead (?=.*) will always match and is completely pointless. Change it to (.*) if you just want to capture that part of the string.

I think the problem is your positive lookahead on(?!\-\w) is trying to match any on that is not followed by - then \w. I think what you want instead is on(?!\-|\w), which matches on that is not followed by - OR \w