Javascript RegExp find and replace empty square brackets

Question

I have a string that contains something like name="text_field_1[]" and I need to find and replace the '1[]' part so that I can increment like this '2[]' , '3[]' , etc.

Code:

$search = new RegExp('1[]', 'g');
$replace = $number + '[]';
$html = $html.replace($search, $replace)

Answer 1

You can use \\d in your regexp whitch means that onlu numbers used before [] . Also you need to escape [] because of it's special characters in regexp.

$search = new RegExp('\\d+\\[\\]', 'g');
$replace = $number + '[]';
$html = $html.replace($search, $replace)

Code: http://jsfiddle.net/VJYkc/1/

Answer 2

You can use callbacks.

var $counter = 0;

$html = $html.replace(/1\[\]/g, function(){
    ++$counter;
    return $counter+'[]';
});

If you need [] preceded by any number, you can use \\d :

var $counter = 0;
$html = $html.replace(/\d\[\]/g, function(){
    ++$counter;
    return $counter+'[]';
});

Note:

• escape brackets, because they are special in regex.
• be sure that in $html there is only the pattern you need to replace, or it will replace all 1[] .

Answer 3

Braces must be escaped within regexps...

var yourString="text-field_1[]";
var match=yourString.match(/(\d+)\[\]/);
yourString=yourString.replace(match[0], parseInt(match[1]++)+"[]");

Answer 4

Here's something fun. You can pass a function into string.replace .

  var re = /(\d+)(\[\])/g/;
  html = html.replace(re, function(fullMatch, value, braces) {
    return (parseInt(value, 10)+1) + braces;
  }

Now you can replace multiple instances of #[] in your string.