Regex to grab strings between square brackets

Question

I have the following string: pass[1][2011-08-21][total_passes]

How would I extract the items between the square brackets into an array? I tried

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

 var s = 'pass[1][2011-08-21][total_passes]'; var result = s.match(/\\[(.*?)\\]/); console.log(result);

but this only returns [1] .

Not sure how to do this.. Thanks in advance.

Answer 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/

Answer 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]' ]

Answer 3

将全局标志添加到您的正则表达式，并迭代返回的数组。

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

Answer 4

'pass[1][2011-08-21][total_passes]'.match(/\[.+?\]/g); // ["[1]","[2011-08-21]","[total_passes]"]

Explanation

\[       # match the opening [
           Note: \ before [ tells that do NOT consider as a grouping symbol.
   .+?   # Accept one or more character but NOT greedy
\]       # match the closing ] and again do NOT consider as a grouping symbol
/g       # do NOT stop after the first match. Do it for the whole input string.

You can play with other combinations of the regular expression https://regex101.com/r/IYDkNi/1

Answer 5

I'm not sure if you can get this directly into an array. But the following code should work to find all occurences and then process them:

var string = "pass[1][2011-08-21][total_passes]";
var regex = /\[([^\]]*)\]/g;

while (match = regex.exec(string)) {
   alert(match[1]);
}

Please note: i really think you need the character class [^\\]] here. Otherwise in my test the expression would match the hole string because ] is also matches by .*.

Answer 6

[C#]

        string str1 = " pass[1][2011-08-21][total_passes]";
        string matching = @"\[(.*?)\]";
        Regex reg = new Regex(matching);
        MatchCollection matches = reg.Matches(str1);

you can use foreach for matched strings.