Remove a particular part of a url string using javascript

Question

In the following string,

http://blahblah.in/folder1/data?number=1&name= &customerId=123

I'm trying to validate the URL, that is, if any querystring parameter does not have a value, remove that parameter from the URL.

The resultant URL will be

http://blahblah.in/folder1/data?number=1&customerId=123

I tried to solve the above problem with the following code, and I know I have a mistake somewhere - can you help me find it?

var url = "http://blahblah.in/folder1/data?number=1&name= &customerId=123";


var tempQuestion = url.split("?");
 var temp = tempQuestion[1].split("=");
var temp2="";
var leng = temp.length;
for(i=0;i<leng;i++){


    if(temp[i].charAt(0)!="&")
        temp2 = temp2.concat(temp[i],"=");
    else 
    {
        temp2 = temp2.concat(temp[i],"");
        var x = temp2.split("&");
        temp2 = temp2.concat(x[0],"=");
    }

}

var result= tempQuestion[0].concat("?",temp2);

mplungjan · Accepted Answer · 2016-06-30 15:05:03Z

3

Just use regular expressions as follows:

    var url = 'http://blahblah.in/folder1/data?number=1&name=&other=&something=&customerId=123'
    url = url.replace(/[a-z]*=&/g,"");
    console.log(url)
    // result is: http://blahblah.in/folder1/data?number=1&customerId=123

edited Jun 30, 2016 at 15:05

mplungjan

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answered Jun 30, 2016 at 15:01

leo.fcx

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3 Comments

SArnab Over a year ago

You should add a case-insensitive flag as well. This will also not work for trailing empty query parameters. But still a very good answer based on OP's example and request.

Aaron Over a year ago

Query parameters name can contain more than lowercase and uppercase letters. I would recommend using the negated character class [^&=] since AFAIK only those characters have any special meaning in this context.

Kishore L Over a year ago

Thank you... you are a life saver.

Iso · Accepted Answer · 2016-06-30 15:07:36Z

1

I’d do it this way:

url.replace(/\?.*/, function(qs) {
  return qs.split('&').filter(function(parameterTuple) {
    return parameterTuple.split('=')[1];
  }).join('&');
});

answered Jun 30, 2016 at 15:07

Iso

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2 Comments

mplungjan Over a year ago

url.replace does not store the value. You need to do url=url.replace(...)

Kishore L Over a year ago

Thank you for your response.. i'll try this method also

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