Simple Javascript replace with a loop

Question

I'm try to replace all occurances wihtin a string with the array index value as below.

var str = '<a href="{0}" title="{1}">{1}</a>';
var params= [];
params.push('Url', 'TitleDisplay');

for (i in params) {
    var x = /'{' + i + '}'/g;
    str = str.replace(x, params[i]);
}

No matter what I do, I cannot seem to get it to work. Dropping the '/g' works with one match, but not all. I know this is basic but for the lide of me I cannot get it to work.

Don't use `for... in... ` for arrays – Andreas Jul 12 '12 at 06:35 — Andreas, Jul 12 '12 at 06:35

Christophe · Accepted Answer · 2012-07-12T17:06:12.917

8

Fiddle here

Code:

var rx = /{([0-9]+)}/g;
str=str.replace(rx,function($0,$1){return params[$1];});

The replace method loops through the string (because of /g in the regex) and finds all instances of {n} where n is a number. $1 captures the number and the function replaces {n} with params[n].

edited Jul 12 '12 at 17:06

answered Jul 12 '12 at 07:44

Christophe

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score 2 · Answer 2 · answered Jul 12 '12 at 06:36

2

try using this:

var x = new RegExp("\\{" + i + "\\}", "g");

instead of this:

var x = /'{' + i + '}'/g;

answered Jul 12 '12 at 06:36

GottZ

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score 2 · Answer 3 · answered Jul 12 '12 at 06:36

You can build a regexp object if you need it to be dynamic

var str = '<a href="{0}" title="{1}">{1}</a>';
var params= [];
params.push('Url', 'TitleDisplay');

for (var i = 0; i < params.length; i++) {
    var x = new RegExp('(\\{'+i+'\\})', 'g');
    str = str.replace(x, params[i]);
}
alert(str);

http://jsfiddle.net/LByBT/

score 1 · Answer 4 · answered Aug 07 '14 at 19:29

1

How about this if you would like to skip a regex solution ..

function replaceAllOccurrences(inputString, oldStr, newStr) 
{
    while (inputString.indexOf(oldStr) >= 0)
    {
        inputString = inputString.replace(oldStr, newStr);
    }

    return inputString;
}

answered Aug 07 '14 at 19:29

Gene Bo

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Can you share your code and the input strings you use - for me I use this successfully across a variety of projects – Gene Bo Jul 06 '15 at 01:32
@emotality that is because probably you are replacing a string with the same string with something in front of or at the end of it – Vasil Gerginski Feb 05 '21 at 20:05

Simple Javascript replace with a loop

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