javascript - Removing all instances of a character from a string

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?

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?

Share Improve this question edited Jul 24, 2014 at 11:46 asked Jul 24, 2014 at 11:24 DGibbs 14.6k7 gold badges49 silver badges88 bronze badges

Add a ment |

5 Answers 5

Sorted by: Reset to default 11

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,'');

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.

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

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

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, '')

科技改变生活-雨落星辰 - 所有的伟大,都源于一个勇敢的开始

javascript - Removing all instances of a character from a string - Stack Overflow

5 Answers 5

与本文相关的文章

评论列表(0)