CODEWITHSUNDEEP
javascript
Vinod
Vinod

Reputation: 533

Query parameter to strip value that contains @ and

Example URL: https://www.example.com/?test=123&[email protected]&xyz=1233 Expected output: https://www.example.com/?test=123&xyz=1233

Basically i am trying to strip the query parameter which containers email ids.

Here is what i have written:

    var url = parent.window.location.href.replace('%40', '@').replace('&', '&');
  if (url.includes('@') && url.includes('.')) {
    var returnurl = '';
    var urlParams = new URLSearchParams(url.substring(url.indexOf('?')));
    var keys = urlParams.keys();
    var entries = urlParams.entries();
    for (pair of entries) {
      if (!(pair[1].includes('@') && pair[1].includes('.'))) {
        if (returnurl != '') {
          returnurl = returnurl + '&' + pair[0] + '=' + pair[1];
        } else {
          returnurl = pair[0] + '=' + pair[1];
        }
      } else {
        url = url.replace(pair[0] + '=' + pair[1], '')
      }
    }
  }
url = url.replace('&&','&');

Is there any better way to manage this? I am looking to write it in pure javascript, so need to remove ECMASCRIPT6 related code(for loop in above sample).

Upvotes: 0

Views: 77

Answers (2)

Dhananjai Pai
Dhananjai Pai

Reputation: 6015

var url = "https://www.example.com/?test=123&[email protected]&xyz=1233"

var path = url.split('?')[0];
var qp = url.split('?')[1];
qp = qp.split('&').filter(function(param) { return param.indexOf('@') === -1} ).join('&');

var result = path+'?'+qp;

console.log(result);

Upvotes: 4

Ori Drori
Ori Drori

Reputation: 191976

You can use the URL constructor, iterate the URLSearchParams instance with it's forEach method, and delete keys that have the @ sign in their value, and then convert the URL back to string:

function stripEmail(urlStr) {
  var url = new URL(urlStr);
  
  url.searchParams.forEach(function(value, key) {
    if(value.includes('@')) url.searchParams.delete(key);
  });
  
  return url.toString();
}


var url = 'https://www.example.com/?test=123&[email protected]&xyz=1233';

var result = stripEmail(url);

console.log(result);

Upvotes: 2

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