How do I replace an asterisk in Javascript using replace()?

Question

Using JQuery, I'm extracting the value from what is essentially a query box for some data in a MySQL database. Knowing that most users will use an '*' (asterisk) as a wildcard value and that MySQL uses the '%' character, I want to convert any asterisks to '%'.

Normally, this would just be as simple as using

queryString = inputText.replace('*', '%');

but I've had little success with that. Javascript's replace() function only replaces the first occurrence, so it's not replacing the last asterisk in something like *Foo*

The second option I tried was using a regular expression, which also didn't work:

queryString = inputText.replace(/\x2a/g, '%');

How can I replace the '*' (asterisks) with a '%' (percent sign)? I'd imagine there's a really simple regular expression, or something I'm overlooking.

Actually, "*Foo*".replace(/\x2a/g, '%') worked for me. What was the problem with that one? Or which browser did it fail in? — Roatin Marth
– Roatin Marth, Commented Oct 2, 2009 at 14:59
It's highly possible I could have botched something on that one. I just tested it again and it's working - weird. I'm testing in Firefox 3.5 nightlies (Shiretoko) and Epiphany (webkit). — S Pangborn
– S Pangborn, Commented Oct 2, 2009 at 15:08
Also, it could have easily been something that was fixed in the newest nightly. I should have kept the old build around to test in. — S Pangborn
– S Pangborn, Commented Oct 2, 2009 at 15:12

Roatin Marth · Accepted Answer · 2009-10-02 14:56:34Z

42

Try:

queryString = inputText.replace(/\*/g, '%');

answered Oct 2, 2009 at 14:56

Roatin Marth

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1 Comment

user2831723 Over a year ago

So simple, yet so hard. Awesome :)

Ateş Göral · Accepted Answer · 2009-10-02 14:58:16Z

5

Splitting a string into an array and then joining it back into a string is faster than regular expression replacements:

queryString = inputText.split("*").join("%");

answered Oct 2, 2009 at 14:58

Ateş Göral

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3 Comments

S Pangborn Over a year ago

I've never seen this approach, but I'll certainly do some of my own testing and try it out. Thanks!

PhiLho Over a year ago

Interesting. Would need some benchmark, both on CPU and memory... Unless you do replacement on a large text, I doubt we see much speed gain.

Roatin Marth Over a year ago

Here's an explanation I found: stackoverflow.com/questions/441018/…

TorNadO · Accepted Answer · 2020-12-04 13:41:30Z

1

If you want to try passwords with a nice touch for remembering

str.replace(str.substr(1,str.length-3), str.substr(1,str.length-3).replace(/./g,"*"))

The output will be

1*********xY

answered Dec 4, 2020 at 13:41

TorNadO

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Comments

Mohammad Maaz · Accepted Answer · 2022-10-20 10:07:47Z

1

Actually, you don't need a regular expression. You can simply use replaceAll:

queryString = inputText.replaceAll("*", "%");

answered Oct 20, 2022 at 10:07

Mohammad Maaz

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Pang · Accepted Answer · 2016-01-22 02:39:11Z

0

You can use:

queryString = inputText.replace(/[*]/g,"");

edited Jan 22, 2016 at 2:39

Pang

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answered Jul 5, 2012 at 6:34

Milos Cuculovic

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Comments

Marcos Villavicencio · Accepted Answer · 2017-09-18 18:50:38Z

0

I use this function to hide some digits of a phone numbers

function replacePhoneNumbers(number){
    var cuantos = number.length;
    var tele = [];
    for (var i = 0; i < cuantos; i++) {
        if(i < 7){
            tele[i] = number[i].replace(/^\d+$/, "*");
        }
        else{
            tele[i] = number[i];
        }
    }
    
    var full_phone = "";
    for (var i = 0; i < cuantos; i++) {
        full_phone += tele[i];
    }
    return full_phone;
    //return number;
}

var number = replacePhoneNumbers("7533930048");
document.getElementById("demo").innerHTML = number;

<p id="demo"></p>

answered Sep 18, 2017 at 18:50

Marcos Villavicencio

1

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