Remove querystring from URL

问题:

What is an easy way to remove the querystring from a Path in Javascript? I have seen a plugin for Jquery that uses window.location.search. I can not do that: The URL in my case is a variable that is set from AJAX.

var testURL = \'/Products/List?SortDirection=dsc&Sort=price&Page=3&Page2=3&SortOrder=dsc\'

回答1:

An easy way to get this is:

function getPathFromUrl(url) {
  return url.split(\"?\")[0];
}

For those who also wish to remove the hash (not part of the original question) when no querystring exists, that requires a little bit more:

function stripQueryStringAndHashFromPath(url) {
  return url.split(\"?\")[0].split(\"#\")[0];
}

EDIT

@caub (originally @crl) suggested a simpler combo that works for both query string and hash (though it uses RegExp, in case anyone has a problem with that):

function getPathFromUrl(url) {
  return url.split(/[?#]/)[0];
}

回答2:

2nd Update: In attempt to provide a comprehensive answer, I am benchmarking the three methods proposed in the various answers.

var testURL = \'/Products/List?SortDirection=dsc&Sort=price&Page=3&Page2=3\';
var i;

// Testing the substring method
i = 0;
console.time(\'10k substring\');
while (i < 10000) {
    testURL.substring(0, testURL.indexOf(\'?\'));
    i++;
}
console.timeEnd(\'10k substring\');

// Testing the split method
i = 0;
console.time(\'10k split\');
while (i < 10000) {
    testURL.split(\'?\')[0]; 
    i++;
}
console.timeEnd(\'10k split\');

// Testing the RegEx method
i = 0;
var re = new RegExp(\"[^?]+\");
console.time(\'10k regex\');
while (i < 10000) {
    testURL.match(re)[0]; 
    i++;
}
console.timeEnd(\'10k regex\');

Results in Firefox 3.5.8 on Mac OS X 10.6.2:

10k substring:  16ms
10k split:      25ms
10k regex:      44ms

Results in Chrome 5.0.307.11 on Mac OS X 10.6.2:

10k substring:  14ms
10k split:      20ms
10k regex:      15ms

Note that the substring method is inferior in functionality as it returns a blank string if the URL does not contain a querystring. The other two methods would return the full URL, as expected. However it is interesting to note that the substring method is the fastest, especially in Firefox.

---

1st UPDATE: Actually the split() method suggested by Robusto is a better solution that the one I suggested earlier, since it will work even when there is no querystring:

var testURL = \'/Products/List?SortDirection=dsc&Sort=price&Page=3&Page2=3\';
testURL.split(\'?\')[0];    // Returns: \"/Products/List\"

var testURL2 = \'/Products/List\';
testURL2.split(\'?\')[0];    // Returns: \"/Products/List\"

---

Original Answer:

var testURL = \'/Products/List?SortDirection=dsc&Sort=price&Page=3&Page2=3\';
testURL.substring(0, testURL.indexOf(\'?\'));    // Returns: \"/Products/List\"

回答3:

A simple way is you can do as follows

public static String stripQueryStringAndHashFromPath(String uri) {
 return uri.replaceAll((\"(\\\\?.*|\\\\#.*)\"), \"\");
}

回答4:

If you\'re into RegEx....

var newURL = testURL.match(new RegExp(\"[^?]+\"))

回答5:

var path = \"path/to/myfile.png?foo=bar#hash\";

console.log(
    path.replace(/(\\?.*)|(#.*)/g, \"\")
);

回答6:

This may be an old question but I have tried this method to remove query params. Seems to work smoothly for me as I needed a reload as well combined with removing of query params.

window.location.href = window.location.origin + window.location.pathname;

Also since I am using simple string addition operation I am guessing the performance will be good. But Still worth comparing with snippets in this answer

回答7:

If you need to perform complex operation on URL, you can take a look to the jQuery url parser plugin.

回答8:

If using backbone.js (which contains url anchor as route), url query string may appear:

1. before url anchor:

   var url = \'http://example.com?a=1&b=3#routepath/subpath\';
   

2. after url anchor:

   var url = \'http://example.com#routepath/subpath?a=1&b=3\';
   

Solution:

window.location.href.replace(window.location.search, \'\');
// run as: \'http://example.com#routepath/subpath?a=1&b=3\'.replace(\'?a=1&b=3\', \'\');