无法匹配除 ' 和 " 之外的任何字符的正则表达式

Question

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 ie , 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}} (eg {{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.

Answer 1

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);

Answer 2

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.