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passing multiple values using xmlhttp.send

Posted on 2010-08-21
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2,496 Views
Last Modified: 2012-05-10
I am using the script below to pass values from a form to a php page.  The first xmlhttp.send always returns a value, the second one is always blank. I am trying to figure out how to send multiple values using xmlhttp.send
if(xmlhttp) { 
  	var txtname = document.getElementById("txtname");
	var number  = document.getElementById("number");
    xmlhttp.open("POST","testing.php",true); //calling testing.php using POST method
    xmlhttp.onreadystatechange  = handleServerResponse;
    xmlhttp.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');
    xmlhttp.send("txtname=" + txtname.value); //Posting txtname to PHP File
	xmlhttp.send("number=" + number.value); //Posting number to PHP File
  }

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Question by:brad0525
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4 Comments
 
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leakim971 earned 500 total points
ID: 33492405
what about :


xmlhttp.send("txtname=" + txtname.value + "&number=" + number.value);

or

xmlhttp.send("txtname=" + encodeURI(txtname.value) + "&number=" + encodeURI(number.value));

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Author Closing Comment

by:brad0525
ID: 33492416
Thanks works perfect
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LVL 1

Expert Comment

by:FractalPat
ID: 33492420
Have you tried something like this:

xmlHttp.send("txtname=txtname.value&number=number.value");
0
 
LVL 82

Expert Comment

by:leakim971
ID: 33492421
You're welcome! Thanks for the points!
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