CodingForums.com - View Single Post - Passing PHP variable to Javascript using AJAX

sunfighter · 02-23-2013, 02:41 PM

In your html file this line

Code:

xmlhttp.open("GET","2.php?q=",true);

is not sending anything as q... So your php query looks like this

PHP Code:


			
SELECT * FROM step1 WHERE id = ''

.
Fix that first.

Then in your loop

Code:

for(i=1; i<="2.php?q="; i++)

is totally meaningless.

The number of rows sent back by the php is only being written into a div with an id of myDiv. I think you want that number in the loop commands.

Take this line out

Code:

<body onload="fone()">

and use it instead of the <body> tag you have. Plus fix up your html.

If your php sent back 2, and you can force this by making your php file be:

PHP Code:


			
echo 2;

And you use this html:

Code:

<!DOCTYPE html>
<html>
<head>
<title>Title of the document</title>
</head>
<body onload="fone()">
<div id="myDiv">div</div>

<script>
function fone()
{
var xmlhttp;
xmlhttp=new XMLHttpRequest();
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
	document.getElementById("myDiv").innerHTML=xmlhttp.responseText;
	myFunction(xmlhttp.responseText);
}
}
xmlhttp.open("GET","2.php?q=",true);
xmlhttp.send('');


}

function myFunction(y)
{
	for(i=1; i<=y; i++)
	{
		var btn=document.createElement("BUTTON");
		btn.id = "join_btn" + i;
		btn.onclick = function() {
			var getID = parseInt(this.id.split("join_btn")[1]);
			document.location.href = pages[getID-1];
		}
		var t=document.createTextNode("JOIN");
		document.body.appendChild(btn);
		btn.appendChild(t);
		document.body.appendChild(document.createElement("br"));
		document.body.appendChild(document.createElement("br"));
	}
}

</script>
</body>
</html>

Things should work. Then figure out what you want to send as q and try it again.