how to send a url using Javascript ajax?

Question

Possible Duplicate:
How to encode a URL in JavaScript?

I am trying to send a url using the following code to a php code, but as the url include &a=12&b=4 once I get the value of the "a" variable in my php code the last part of address is removed.

url = http://www.example.com/help.jpg?x=10&a=12&b=4 but the url that I get in my php file is http://www.example.com/help.jpg?x=10 (&a=12&b=4 is removed, I know the reason is that javascript,ajax mix it up with the url address and do not know its just a value but do not know how to solve it)

         function upload(url){

            if (window.XMLHttpRequest)
            {// code for IE7+, Firefox, Chrome, Opera, Safari
                xmlhttp=new XMLHttpRequest();
            }
            else
            {// code for IE6, IE5
                xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
            }
            xmlhttp.onreadystatechange=function()
            {
                if (xmlhttp.readyState==4 && xmlhttp.status==200)
                {
                    document.getElementById("output").innerHTML= xmlhttp.responseText;
                }
            }
            xmlhttp.open("GET","Photos.php?a="+url,true);
            xmlhttp.send();
     }        


   if(isset($_GET["a"]))
   {
       $Address = $_GET["a"];
       echo $Address;

   }

output is >>> " http://www.example.com/help.jpg?x=10" but it should be http://www.example.com/help.jpg?x=10&a=12&b=4

Check this out: http://stackoverflow.com/questions/332872/how-to-encode-a-url-in-javascript — John V., Jan 07 '13 at 21:33
Closely related: http://stackoverflow.com/questions/332872/how-to-encode-a-url-in-javascript — gd1, Jan 07 '13 at 21:33
As a comment I would say that if you're posting something you should use the POST http method, not GET. Considering your url problem I guess the solution relies into using javascript `url_encode` methods combined with `$_REQUEST` array in php. — Sebas, Jan 07 '13 at 21:35

score 5 · Accepted Answer · answered Jan 07 '13 at 21:34

5

You need to encode the parameter

xmlhttp.open("GET","Photos.php?a="+encodeURIComponent(url),true);

answered Jan 07 '13 at 21:34

Musa

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score 0 · Answer 2 · edited May 16 '14 at 20:27

0

You need to encode the URL. You can use encodeURIComponent(str) and encodeURI(str) like:

var eurl = encodeURIComponent("http://www.example.com/help.jpg?x=10&a=12&b=4");
xmlhttp.open("GET","Photos.php?a=" + eurl, true);

edited May 16 '14 at 20:27

AliciaBytes

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answered Jan 07 '13 at 21:36

bPratik

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how to send a url using Javascript ajax?

2 Answers2