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使用变量时,在Regex中转义方括号?

[英]Escaping square brackets in Regex when using a variable?

I have this code to replace all opening and closing square brackets which has a matching variable inside: 我有这个代码来替换所有打开和关闭方括号,里面有一个匹配的变量:

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'); line var re = new RegExp("["+j+"]", 'g'); doesn't work because I assume the brackets aren't being escaped. 不起作用,因为我认为括号没有被转义。 Does anyone know how I would escape them, but still be able to have a variable in the pattern? 有谁知道我会如何逃避它们,但仍然能够在模式中有一个变量? Thanks! 谢谢! :) :)

你应该用反斜杠来逃避它:

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

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