This is my js code:
html = html.replace("/["+increment+"]/gi", '[' + counter + ']');
where increment is 0 and counter is 1 or
html = html.replace("/[0]/gi", '[1]');
My version does not replace the [0] with [1] in my string. Why ?
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2 Answers
You need to use the RegExp constructor as the regex is dynamic
var regex = new RegExp("\\[" + increment + "\\]", 'gi')
html = html.replace(regex, '[' + counter + ']');
Also you could sanitize the dynamic variable if you want
if (!RegExp.escape) {
//A escape function to sanitize special characters in the regex
RegExp.escape = function (value) {
return value.replace(/[\-\[\]{}()*+?.,\\\^$|#\s]/g, "\\$&")
};
}
//You could also escape the dynamic value it is an user input
var regex = new RegExp("\\[" + RegExp.escape(increment) + "\\]", 'gi')
html = html.replace(regex, '[' + counter + ']');
2 Comments
Jozsef Naghi
ok, but what about this format "@[0], 0" into "@[1], 1" ? thx
Arun P Johny
@JozsefNaghi
var regex = new RegExp('@\\[' + increment + '\\],\s' + increment, 'gi') html = html.replace(regex, '@[' + counter + '], ' + counter);Use this way:
html = html.replace(new RegExp("\\["+increment+"\\]", "gi"), '[' + counter + ']');
This makes use of the dynamic values.
answered Jun 2, 2015 at 12:13
Praveen Kumar Purushothaman
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