Get array of all strings between two characters (Regex + JavaScript)

Question

I'm not expert in Regex and I want just do this thing. Create this array :

["[1+5]", "1+5", "[2*[1+1]+5]", "2*[1+1]+5", "[1+1]", "1+1"]

from this string :

"blalbla[1+5] blabla [2*[1+1]+5] blalbla"

I have tested a lot of methods, I have a headache... Here the link to test what I have did: https://jsfiddle.net/a47e60pd/1/

Thanks a lot :)

If you have arbitrary level of nested brackets, you cannot use a regex. JS regex does not support recursion. Or, you may want to use XRegExp.matchRecursive. — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Aug 28, 2016 at 20:22
What's the point of this exercise? Are you going to evaluate those, ie. replace 1+1 with 2? — georg
– georg, Commented Aug 28, 2016 at 20:27
@georg Yes, then I use eval method. Oriol's answer is satisfactory answer. — Saad
– Saad, Commented Aug 29, 2016 at 10:25

Oriol · Accepted Answer · 2016-08-28 20:37:13Z

2

Don't use regex. Just use a loop.

function* parse(str) {
  var indices = [];
  for(var i=0; i<str.length; ++i) {
    if(str[i] === '[') {
      indices.push(i);
    } else if(str[i] === ']') {
      let idx = indices.pop();
      yield str.slice(idx+1, i);
      yield str.slice(idx, i+1);
    }
  }
}
console.log([...parse("blalbla[1+5] blabla [2*[1+1]+5] blalbla")]);

answered Aug 28, 2016 at 20:37

Oriol

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ABE · Accepted Answer · 2016-08-28 21:36:39Z

1

Is possible with clean regex like this:

/\[[^[]+\]|\[.+?\]([^[])+\]/g

Explanation: Group 1: \[[^[]+\] find brackets that start and finish without other brackets.

Or | Group 2:

\[.+?\] find the start of brackets, until the finish brackets, and ([^[])+\] continue find all characters that not "[" until the last "]"

var ExpToReturn;

var text = "blalbla[1+5] blabla [2*[1+1]+5] blalbla";

var str = text.match(/\[[^[]+\]|\[.+?\]([^[])+\]/g);

ExpToReturn = (str);

alert (ExpToReturn);

edited Aug 28, 2016 at 21:36

answered Aug 28, 2016 at 21:23

ABE

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1 Comment

Saad Over a year ago

Your method is also good. But I prefer Oriol's method. Thanks a lot for your useful answer

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