Javascript nested square brackets in string

Question

I am looking for an easier (and less hacky) way to get the substring of what is inside matching square brackets in a string. For example, lets say this is the string:

[ABC[D][E[FG]]HIJK[LMN]]OPQR[STUVW]XYZ

I want the substring:

ABC[D][E[FG]]HIJK[LMN]

Right now, I am looping through the string and counting the open and closed brackets, and when those numbers are the same, I take substring of the first open bracket and last closed bracket.

Is there an easier way to do this (ie with regex), so that I do need to loop through every character?

You're parsing a little language, with nesting. There is nothing hacky whatsoever about iterating over the characters in the input--that's what parsers do. What would be hacky is a hard-to-read, hard-to-maintain, limited regexp such as was suggested in one answer. — user663031
– user663031, Commented Mar 21, 2015 at 18:45
@torazaburo Hey, thanks, I think your right; I am just going to leave it as is. I just thought that if there was an easier and clean way with regex, I could use that instead. — Matthew S.
– Matthew S., Commented Mar 21, 2015 at 19:36

user663031 · Accepted Answer · 2015-03-22 03:44:12Z

1

Here's another approach, an ugly hack which turns the input into a JS array representation and then parses it using JSON.parse:

function parse(str) {
    return JSON.parse('[' +
        str.split('') . join(',') .  // insert commas
            replace(/\[,/g, '[')  .  // clean up leading commas
            replace(/,]/g, ']')   .  // clean up trailing commas
            replace(/\w/g, '"$&"')   // quote strings
        + ']');
}

>> hack('A[B]C')
<< ["A", ["B"], "C"]

Now a stringifier to turn arrays back into the bracketed form:

function stringify(array) {
    return Array.isArray(array) ? '[' + array.map(stringify).join('') + ']' : array;
}

Now your problem can be solved by:

stringify(parse("[ABC[D][E[FG]]HIJK[LMN]]OPQR[STUVW]XYZ")[0])

answered Mar 22, 2015 at 3:44

user663031

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Comments

Alhamza Alnaimi · Accepted Answer · 2015-03-21 18:30:44Z

0

Not sure if I get the question right (sorry about that).

So you mean that if you were to have a string of characters X, you would like to check if the string combination Y is contained within X?

Where Y being ABC[D][E[FG]]HIJK[LMN]

If so then you could simply do:

var str = "[ABC[D][E[FG]]HIJK[LMN]]OPQR[STUVW]XYZ"; var res = str.match(/ABC\[D]\[E\[FG]]HIJK\[LMN]/);

The above would then return the string literal Y as it matches what is inside str.

It is important that you pay attention to the fact that the symbols [ are being escaped with a \. This is because in regex if you were to have the two square brackets with any letter in between (ie. [asd]) regex would then match the single characters included in the specified set.

You can test the regex here: https://regex101.com/r/zK3vZ3/1

answered Mar 21, 2015 at 18:30

Alhamza Alnaimi

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1 Comment

vp_arth Over a year ago

Aha, so literally, why not just res = 'ABC[D][E[FG]]HIJK[LMN]'?

Wiktor Stribiżew · Accepted Answer · 2016-04-04 08:18:01Z

I think the problem is to get all characters from an opening square bracket up to the corresponding closing square bracket. Balancing groups are not implemented in JavaScript, but there is a workaround: we can use several optional groups between these square brackets.

The following regex will match up to 3 nested [...] groups and you can add the capturing groups to support more:

\[[^\]\[]*(?: \[[^\]\[]*(?: \[[^\]\[]*(?:\[[^\]\[]*\])*\] )*[^\]\[]* \][^\]\[]* )*[^\]\[]* \]

See example here. However, performance may be not that high with such heavy backtracking.

UPDATE

Use XRegExp:

var str = '[ABC[D][E[FG]]HIJK[LMN]]OPQR[STUVW]XYZ';

// First match:
var res = XRegExp.matchRecursive(str, '\\[', ']');
document.body.innerHTML = "Getting the first match:<br/><pre>" + JSON.stringify(res, 0, 4) + "</pre><br/>And now, multiple matches (add \"g\" modifier when defining the XRegExp)";

// Multiple matches:
res = XRegExp.matchRecursive(str, '\\[', ']', 'g');
document.body.innerHTML += "<pre>" + JSON.stringify(res, 0, 4) + "</pre>";

<script src="https://cdnjs.cloudflare.com/ajax/libs/xregexp/2.0.0/xregexp-all-min.js"></script>

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