var name = document.getElementById('name');

farkewie · December 11, 2007

Hello,

new to Javascipt, i hace downloaded the yahoo ajax tools and i am trying to get some post requests working without loading the page,

i have tried to add some code so it sends a username and password from text fields but its not working here is what i get when i do print_r($_POST)

Array (

[name] => [object]

[pass] => [object]

)

and here is my code - this is just testing once i get this working ill start coding allthe page properly.


<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>

<head>
<meta http-equiv="content-type" content="text/html; charset=iso-8859-1">
<meta name="author" content="fuckme">
<script src="js/yahoo.js"></script>
<script src="js/event.js"></script>
<script src="js/connection.js"></script>
<title>Untitled 9</title>
</head>

<body>
<form name="form1" method="post" action="" id="form1">
<table width="204">
<tr><td>name</td><td>pass</td>
</tr>
    <tr>
      <td><label>
        <input type="text" name="name" id="name">
      </label></td>
      <td><label>
        <input type="text" name="pass" id="pass">
      </label></td>
    </tr>
  </table>
  <label>
<input type="button" name="button" id="button" value="Submit" onClick="makeRequest();" >
  </label>
</form>


<div id="container"></div>

<script>
var div = document.getElementById('container');

var handleSuccess = function(o){
YAHOO.log("The success handler was called.  tId: " + o.tId + ".", "info", "example");
if(o.responseText !== undefined){
	div.innerHTML +=  o.responseText;

}
};

var handleFailure = function(o){
	YAHOO.log("The failure handler was called.  tId: " + o.tId + ".", "info", "example");

if(o.responseText !== undefined){
	div.innerHTML = "<li>Transaction id: " + o.tId + "</li>";
	div.innerHTML += "<li>HTTP status: " + o.status + "</li>";
	div.innerHTML += "<li>Status code message: " + o.statusText + "</li>";
}
};

var callback =
{
  success:handleSuccess,
  failure:handleFailure,
  argument:['foo','bar']
};
var name = document.getElementById('name');
var pass = document.getElementById('pass');
YAHOO.util.Connect.setForm(form1); 
var sUrl = "post.php";
var postData = "name=" + name +"&pass=" + pass +"";

function makeRequest(){

var request = YAHOO.util.Connect.asyncRequest('POST', sUrl, callback, postData);

YAHOO.log("Initiating request; tId: " + request.tId + ".", "info", "example");

}

YAHOO.log("As you interact with this example, relevant steps in the process will be logged here.", "info", "example");
</script>





</body>
</html>

Thank you

phpQuestioner · December 11, 2007

here's a good ajax form tutorial - take a look at it and go from there:

http://www.tizag.com/ajaxTutorial/ajaxform.php

I did notice your not including your post url in you ajax function - I think it should be like this:

<input type="button" name="button" id="button" value="Submit" onClick="makeRequest('process-login.php');" >

also ajax questions belong on this forum:

http://www.phpfreaks.com/forums/index.php/board,51.0.html

this forum is just for JavaScript - Good Luck.

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