Removing all instances of a character from a string

Question

I'm attempting to remove all instances of a given set of characters £$€,. from a string in jQuery/Javascript. I'm using the replace function, however this only appears to remove a single instance of the character and not all of them.

For example consider the string:

1,500,00.00.$djdjd£10€10

I get back:

1500,0000.djdjd1010

As you can see, it only removes a single instance of each character. £, $ and € are fine as there is only one of each in the string.

Here is what I have so far:

function validatePriceRange(value, min, max) {

    var replacements = ["£", "$", "€", ",", "."];

    $.each(replacements, function (index, item) {
        value = value.replace(item, "");
    });

    var value = parseInt(value, 10);

    return value >= min && value <= max;
}

jsFiddle

Can anyone spot what I've done wrong?

score 9 · Accepted Answer · answered Jul 24 '14 at 11:28

9

replace called with a string as first argument does only one replacement, while using a regular expression with flag g replaces all occurrences.

Using a regular expression, you can also avoid looping over an array and do it in one pass :

value = value.replace(/£|\$|€|,|\./g,'');

answered Jul 24 '14 at 11:28

Denys Séguret

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Perfect - thanks for the explanation on how replace works. I ignorantly assumed it worked in a similar way to how .`Replace()` works in `C#`. – DGibbs Jul 24 '14 at 11:40

score 1 · Answer 2 · edited May 23 '17 at 11:54

1

You are only cycling through your replacement array once and replace everytime the specific character. But replace is only replacing the first occurance of a given string.

For a replace all method, look here.

edited May 23 '17 at 11:54

Community

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answered Jul 24 '14 at 11:31

Florian Leitgeb

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yay, pointing at existing solutions rocks. I might consider extending that solution by making ReplaceAll a prototype of String. – gillyspy Jul 24 '14 at 11:41

KooiInc · Answer 3 · 2014-07-24T11:48:39.077

I don't think you need a function for it:

var validated = parseInt('1,500,00.00.$djdjd£10€10'.replace(/[£$€,.]/g,''), 10);
//=> 15000000
// or if you want the validated directly
var validated = function(min,max) {
                  var v = parseInt('1,500,00.00.$djdjd£10€10'
                                    .replace(/[£$€,.]/g,''), 10);
                  return v >= min && v <==max;
                }(1000, 200000); //=> false

The regular expression should be different if you want to include all digits in the string:

var validated = function(min,max) {
                  var v = parseInt('1,500,00.00.$djdjd£10€10'
                                    .replace(/[^\d]/g,''), 10);
                  //                           ^ replace non numbers
                  // v now is 150000001010
                  return v >= min && v <==max;
                }(1000, 200000); //=> false

score 0 · Answer 4 · answered Jul 24 '14 at 11:27

0

Use a regex with the global flag, which will search and replace all instances

var replacements = ["£", "\\$", "€", ",", "\\."];
$.each(replacements, function (index, item) {
    value = value.replace(new RegExp(item, "g"), '');
});

Demo: Fiddle

answered Jul 24 '14 at 11:27

Arun P Johny

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score 0 · Answer 5 · edited May 23 '17 at 12:10

0

As already answered here you could use the following regex that replaces all non characters and whitespaces with empty.

var value = "1,500,00.00.$djdjd£10€10"
value = value.replace(/[^\w\s]/gi, '')

edited May 23 '17 at 12:10

Community

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answered Jul 24 '14 at 11:39

Jarno Lonardi

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Removing all instances of a character from a string

5 Answers5