Regex matching curly bracket

Question

Looking for help in matching the curly brackets in a regular expression pattern. I've tried different combinations of escapes, and symbol matching with little luck. Perhaps because it's Friday afternoon and I'm overlooking something; but your ideas would be greatly appreciated. The code below:

function stringFormat(str, arr) {
   for (var i = 0; i < arr.length; i++) {
        var regExp = new RegExp('^\{' + i + '\}$', 'g');
        str = str.replace(regExp, arr[i]);   
    }
    return str;  
}

var str = '<p>The quick {0}, brown {1}</p>';

$('#test').html(stringFormat(str, ['brown', 'fox']));

I've also started a fiddle on this, http://jsfiddle.net/rgy3y/1/

Answer 1

Instead of trying to match a bunch of different numbers, why not just do it all in one fell swoop:

function stringFormat(str, arr) {
  return str.replace(
      /\{([0-9]+)\}/g,
      function (_, index) { return arr[index]; });
}

On your example,

var str = '<p>The quick {0}, brown {1}</p>';

// Alerts <p>The quick brown, brown fox</p>
alert(stringFormat(str, ['brown', 'fox']));

This has the benefit that nothing weird will happen if arr contains a string like '{1}'. E.g.
stringFormat('{0}', ['{1}', 'foo']) === '{1}' consistently instead of 'foo' as with the fixed version of the original, but inconsistently with stringFormat('{1}', ['foo', '{0}']) === '{0}'

Answer 2

To get a \ in a string literal you need to type \\. In particular, '\{' == '{'. You want '\\{'.