javascript - Escaping square brackets in Regex when using a variable? - Stack Overflow

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I have this code to replace all opening and closing square brackets which has a matching variable inside:

for (var j = 0; j <= temp.length; j++) {
    var re = new RegExp("["+j+"]", 'g');
    imgData = imgData.replace(re, temp[j]);
}

The line var re = new RegExp("["+j+"]", 'g'); doesn't work because I assume the brackets aren't being escaped. Does anyone know how I would escape them, but still be able to have a variable in the pattern? Thanks! :)

I have this code to replace all opening and closing square brackets which has a matching variable inside:

for (var j = 0; j <= temp.length; j++) {
    var re = new RegExp("["+j+"]", 'g');
    imgData = imgData.replace(re, temp[j]);
}

The line var re = new RegExp("["+j+"]", 'g'); doesn't work because I assume the brackets aren't being escaped. Does anyone know how I would escape them, but still be able to have a variable in the pattern? Thanks! :)

• javascript
• regex
• replace

Share Improve this question Follow asked Nov 29, 2012 at 1:43 Joey MoraniJoey Morani 26.6k3333 gold badges8989 silver badges132132 bronze badges

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1 Answer 1

Sorted by: Reset to default 12

You should escape it with backslashes:

var re = new RegExp("\\[" + j + "\\]", "g");

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