JavaScript-Regular expression matches the contents of the innermost brackets

Now there is a string:

str1 = '(subject_id = "A" OR (status_id = "Open" AND (status_id = "C" OR level_id = "D")))'

Or

str2 = '(subject_id = "A" OR subject_id = "Food" OR (subject_id = "C" OR (status_id = "Open" AND (status_id = "C" OR (level_id = "D" AND subject_id = "(Cat)")))))'

I need to pass the regular, matching the innermost parentheses in the string and their contents (not matching the parentheses in the quotation marks), i.e.:

str1 => (status_id = "C" OR level_id = "D")str2 => (level_id = "D" AND subject_id = "(Cat)")

So what is this hyper-complex regular supposed to write?

If the regular implementation is not, then how to achieve JS?

Add, for str1 , I found such a regular can satisfy the match:

\([^()]+\)

But for str2, there is still no way to look forward to the answer!

Reply content:

Now there is a string:

str1 = '(subject_id = "A" OR (status_id = "Open" AND (status_id = "C" OR level_id = "D")))'

Or

str2 = '(subject_id = "A" OR subject_id = "Food" OR (subject_id = "C" OR (status_id = "Open" AND (status_id = "C" OR (level_id = "D" AND subject_id = "(Cat)")))))'

I need to pass the regular, matching the innermost parentheses in the string and their contents (not matching the parentheses in the quotation marks), i.e.:

str1 => (status_id = "C" OR level_id = "D")str2 => (level_id = "D" AND subject_id = "(Cat)")

So what is this hyper-complex regular supposed to write?

If the regular implementation is not, then how to achieve JS?

Add, for str1 , I found such a regular can satisfy the match:

\([^()]+\)

But for str2, there is still no way to look forward to the answer!

For str2, I found something like this.

\([^()]*\"[^"]*\"[^()]*\)

Looking at the requirements I didn't even think about using the regular, it seems too complicated ... Go directly to the traditional method;
You can use the idea of the arithmetic precedence , that is, the data structure of the stack to get the contents of the inner brackets;
Technical points:

1. Match the parentheses of the inner layer

2. The content within the quotation marks is not used as a matching criterion

Start designing the algorithm according to this idea:
The algorithm is to calculate the substring to match startIndex and endIndex then use the substring() method to obtain the substring;

• When matching to a "(" character, into the stack , when we match to the first ")" , out of the stack , that is, two indexes between the substring as the target string;

• When matched to one "\"" , the match is stopped until the next search is "(" reached and the "\"" search continues to begin "(" .

Beat the brain to think out of the algorithm, there is a lack of welcome to add.

So, try
/\(([^\(\)]*?" [^\"\(\)]*([^\"\(\)]+\)[^\(\)]*?\"[^\(\)]*)+)| ([^\(\)]+\)/

Add:

Analyze Requirements > Find solutions for each demand point > consolidation solution = Problem solving

Analysis Requirements:

1. Need to match ( a ) the form

2. aThere are two possibilities for the characters that are contained in a1 and a2 represent

   1. a1Contains one or more b " c " b forms of a string,

      1. bA string that is not included " , ( or )

      2. Which c is a string that is not included "

   2. a2does not contain ( or)

Inverse derivation:

2.2 a2 =[^\(\)]*
2.1.1 b =[^\(\)\"]*
2.1.2 c =[^\"]*
2.1 = = = = a1 (b\"c\"b)+ (b\"c\")+b([^\(\)\"]*\"[^\"]*\")+[^\(\)\"]*
1 \(a\) = \(a1\)|\(a2\) =\(([^\(\)\"]*\"[^\"]*\")+[^\(\)\"]*\)|\([^\(\)]*\)

Regular Expressions:

/\(([^\(\)\"]*\"[^\"]*\")+[^\(\)\"]*\)|\([^\(\)]*\)/

Verify:

var reg = /\(([^\(\)\"]*\"[^\"]*\")+[^\(\)\"]*\)|\([^\(\)]*\)/;'(the (quick "brown" fox "jumps over, (the) lazy" dog ))'    .match(reg)[0]//"(quick "brown" fox "jumps over, (the) lazy" dog )"'(the ("(quick)" brown fox "jumps (over, the)" lazy) dog )'    .match(reg)[0];//"("(quick)" brown fox "jumps (over, the)" lazy)"'(the (quick brown fox (jumps "over", ((the) "lazy"))) dog )'    .match(reg)[0];//"(the)"

So change it:

substr=str.match(/\([^()]+\)/g)[0]

Get the innermost bracket and the value in it, and then determine whether the previous digit is ", whether the next one is":

index=str.indexOf(str.match(/\([^()]+\)/g)[0])length=str.match(/\([^()]+\)/g)[0].lengthstr.substr(index+length,1)str.substr(index-1,1)

If it does not exist, then it is the answer that is needed, if present, replace the substr in STR first, then in match, and finally in replace:

str.replace(substr,"&&&")str.replace(substr,"&&&").match(/\([^()]+\)/g)[0]str.replace(substr,"&&&").match(/\([^()]+\)/g)[0].replace("&&&",substr)

The difficulty of the problem is to have recursive statistics on "", for example

(level_id = "D AND subject_id = "(Cat)"")

(CAT) is in compliance with the requirements.

\([^()]*?\"((?:[^\"\"]|\"(?1)\")*+)\"[^()]*?\)|\([^()]*?\)

True love of life, away from the regular, the regular can meet your requirements, PHP can be used (PHP support recursion) Java and Python is not available.

Recommend a thought to find (the index, cut string processing

The phone can't send a regular black line
If the "^ ()" In the landlord does not match () then continue
The mismatch (the condition of the removal, the greedy + change to *?)

! code

Console.log (' (subject_id = "A" or (status_id = "Open" and (status_id = "C" or level_id = "D")) '. Match (/(1*)/))
Hope to help you

1. ()

The use of regular matching will be more complex, it is recommended to replace the interference string "(and)", such as "[,]", and then replace with simple regular, and then change back.

The regular Python implementation is as follows:

import restr1 = '(subject_id = "A" OR (status_id = "Open" AND (status_id = "C" OR level_id = "D")))'str2 = '(subject_id = "A" OR subject_id = "Food" OR (subject_id = "C" OR (status_id = "Open" AND (status_id = "C" OR (level_id = "D" AND subject_id = "(Cat)")))))'pat = re.compile(r"""(?<=[^"])        \([^()]+?        ("\(.+?\)")*        \)        (?=[^"])        """, re.X)print pat.search(str1).group(0)print pat.search(str2).group(0)

The output is:

(status_id = "C" OR level_id = "D")(level_id = "D" AND subject_id = "(Cat)")

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