Passing value to php is undefined

[cleary1981]

No the value is assigned in showobject

[acoder]

Yes, but g_objName is set in createObject() on line 197. Check that it's set properly there.

[cleary1981]

It is working fine. I inserted an alert at the end of that function and it gave me the correct result. I also put the same alert at the start of the render function and that worked also.

[cleary1981]

The values in the render funcion are all fine as I used the alerts as shown below and they all prduced the correct results. Does the problem lie in the php code.

function render() {
	alert(g_objName);
	var arr = document.getElementsByTagName("div");
    var data = {
	    'result[]' : [],
    	'xReturnValue[]': [],
    	'yReturnValue[]' : []
    	};
    
    for(var i = 0; i < arr.length; i++)
    	{
    	var xReturnValue = -10;
    	var yReturnValue = -5;
    	for(var elem = arr[i];elem != null;elem = elem.offsetParent)
    		{
    		xReturnValue += elem.offsetLeft;
   			yReturnValue += elem.offsetTop;
 				           
   			}
   			alert(arr[i].innerHTML);
    	data['result[]'].push(arr[i].innerHTML);
    	data['xReturnValue[]'].push(xReturnValue);
    	alert(xReturnValue);
    	data['yReturnValue[]'].push(yReturnValue);
    	alert(yReturnValue);
    	}
    $.post("render.php",data);
    // this bit of code is using jquery 
}

Heres the php code in question

<?php
require "config.php";
foreach($_POST['result'] as $n=>$result){
  $xReturnValue = $_POST['xReturnValue'][$n];
  $yReturnValue = $_POST['yReturnValue'][$n];
  //do what you want with the data
	$q = mysql_query("update object set xpos='$xReturnValue', ypos='$yReturnValue' where object_name = '$result'");
	
}

?>

config.php works fine as I have used it throughout this program

[acoder]

So you're saying arr[i].innerHTML is no longer undefined?

[cleary1981]

Yeah. Dont know whats changed but its working now. lol. Just doesnt update my db. Problem must be in the php

[acoder]

Well, if the values passed are all fine and the call is OK, that just leaves the PHP code. Perhaps you should ask in the PHP forum.

[cleary1981]

OK will try that now. thanks.

[cleary1981]

Hi,

My php seems to be working fine now. I think. All the records are updating except for the very first one. Heres my code now.

Jquery

function render() {
	
	var arr = document.getElementsByTagName("div");
    var data = {
	    'result[]' : [],
    	'xReturnValue[]': [],
    	'yReturnValue[]' : []
    	};
    
    for(var i = 0; i < arr.length; i++)
    	{
    	var xReturnValue = -10;
    	var yReturnValue = -5;
    	for(var elem = arr[i];elem != null;elem = elem.offsetParent)
    		{
    		xReturnValue += elem.offsetLeft;
   			yReturnValue += elem.offsetTop;
 				           
   			}
   			alert(arr[i].innerHTML);
   			var result = arr[i].innerHTML;
    	data['result[]'].push(result);
    	data['xReturnValue[]'].push(xReturnValue);
    	alert(xReturnValue);
    	data['yReturnValue[]'].push(yReturnValue);
    	alert(yReturnValue);
    	}
    $.post("render.php",data);
    // this bit of code is using jquery 
}

php

<?php
require "config.php";


foreach($_POST['result'] as $n=>$result){
   $xReturnValue = $_POST['xReturnValue'][$n];
   $yReturnValue = $_POST['yReturnValue'][$n];
   $q = mysql_query("UPDATE object SET xpos='$xReturnValue', ypos='$yReturnValue' WHERE object_name = '$result'");
}

?>

[acoder]

Does the first div display undefined or is it the name of the first object to be updated?

[cleary1981]

in the alerts it shows the correct information

[acoder]

OK, if you pass these same values in the same format to PHP without Ajax, does it (still not) work?

[cleary1981]

when i do that only random records get updated. I cant see any errors in any of the code. Cud this be a bug in mysql?

[acoder]

The query seems simple enough. How did the thread in the PHP forum go?

[cleary1981]

hadnt much luck with it. Im thinking of rewriting my script all together