Regex to grab strings between square brackets

Solution 1:

You are almost there, you just need a global match (note the /g flag):

match(/\[(.*?)\]/g);

Example: http://jsfiddle.net/kobi/Rbdj4/

If you want something that only captures the group (from MDN):

var s = "pass[1][2011-08-21][total_passes]";
var matches = [];

var pattern = /\[(.*?)\]/g;
var match;
while ((match = pattern.exec(s)) != null)
{
  matches.push(match[1]);
}

Example: http://jsfiddle.net/kobi/6a7XN/

Another option (which I usually prefer), is abusing the replace callback:

var matches = [];
s.replace(/\[(.*?)\]/g, function(g0,g1){matches.push(g1);})

Example: http://jsfiddle.net/kobi/6CEzP/

Solution 2:

var s = 'pass[1][2011-08-21][total_passes]';

r = s.match(/\[([^\]]*)\]/g);

r ; //# =>  [ '[1]', '[2011-08-21]', '[total_passes]' ]

example proving the edge case of unbalanced [];

var s = 'pass[1]]][2011-08-21][total_passes]';

r = s.match(/\[([^\]]*)\]/g);

r; //# =>  [ '[1]', '[2011-08-21]', '[total_passes]' ]

Solution 3:

add the global flag to your regex , and iterate the array returned .

 match(/\[(.*?)\]/g)