jQuery Ajax

How do I know if Ajax is setup on my server?  If it is not, is there a good tutorial that shows you how to on a windows 2008 machine?
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rgranlundAsked:
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Russ SuterSenior Software DeveloperCommented:
AJAX doesn't require any special server setup. It simply relies on existing HTTP GET/POST methods and runs them asynchronously through Javascript calls rather than page requests. Just call the web method using the $.ajax() function.
rgranlundAuthor Commented:
I'm a newbie.  What is wrong with the way I have the following setup?
<html>
<!DOCTYPE >
<html>
<meta charset="utf-8">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
<body>


	<a href="file/ajax.txt">Load AJAX File</a>
	<br /><br />
	
	<a href="file/ajax.txt">Load AJAX HTML</a>
	
	<script src="js/main.js" ></script>
</body>
</html>
</html>

<script>
(function() {
	var link = document.getElementsByTagName("a")[0];
	
	link.onClick = function() {
//  XHR Object
		var xhr = new XMLHttpRequest();
		
//  Handle the On Read State Change Event
	//  xnr. readyState property values
	// 0 = ininitialized
	// 1 = Loading
	// 2 = Loaded
	// 3 = Interactive
	// 4 = Complete
	
		xhr.onreadystatechange = function() {
			if ((xhr.readyState == 4) && (xhr.status == 200 || xhr.status == 304)) {			
				var body = document.getElementByTagName("body")[0];
				var p = document.createElement("p");
				var pText = document.createTextNode(xhr.responseText);
				p.appendChild(pText);
				body.appendChild(p);
			}
		};
	
		
//  Open request
		xhr.open("GET", "files/ajax.text", true);
		
//  Send Request to Server
		xhr.send(null);
		
		return false;
	};

})();

</script>

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When I click the link it goes to the page instead of GETing the info and appending it.
Russ SuterSenior Software DeveloperCommented:
I didn't bother to inspect your code very closely once I realized that you're importing the jQuery library but not using it. jQuery makes AJAX life so much simpler.

What does your server-side method look like? I don't even see where you're referencing the server-side method.
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rgranlundAuthor Commented:
I don't even see where you're referencing the server-side method.?
Not using the jQuery Library?

My  "js/main.js is:
(function() {
	var link = document.getElementsByTagName("a")[0];
	
	link.onClick = function() {
//  XHR Object
		var xhr = new XMLHttpRequest();
		
//  Handle the On Read State Change Event
	//  xnr. readyState property values
	// 0 = ininitialized
	// 1 = Loading
	// 2 = Loaded
	// 3 = Interactive
	// 4 = Complete
	
		xhr.onreadystatechange = function() {
			if ((xhr.readyState == 4) && (xhr.status == 200 || xhr.status == 304)) {			
				var body = document.getElementByTagName("body")[0];
				var p = document.createElement("p");
				var pText = document.createTextNode(xhr.responseText);
				p.appendChild(pText);
				body.appendChild(p);
			}
		};
	
		
//  Open request
		xhr.open("GET", "files/ajax.txt", true);
		
//  Send Request to Server
		xhr.send(null);
		
		return false;
	};

})();

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Russ SuterSenior Software DeveloperCommented:
That's not jQuery. A jQuery AJAX call would look something like this:

  $.ajax({
    url : "files/ajax.txt",
    dataType: "text",
    success : ajaxCallback
    }
  });

function ajaxCallback(result, status, xhr) {
  var body = document.getElementByTagName("body")[0];
  var p = document.createElement("p");
  var pText = document.createTextNode(result.responseText);
  p.appendChild(pText);
  body.appendChild(p);  
}

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rgranlundAuthor Commented:
When I do:
 $.ajax({
    url : "file/ajax.txt",
    dataType: "text",
    success : ajaxCallback
    });
 
 
function ajaxCallback(result, status, xhr) {
	var body = document.getElementByTagName("body")[0];
	var p = document.createElement("p");
	var pText = document.createTextNode(result.responseText);
	p.appendChild(pText);
	body.appendChild(p);
	};

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I get the error:
TypeError: document.getElementByTagName is not a function
      

var body = document.getElementByTagName("body")[0];
Russ SuterSenior Software DeveloperCommented:
Yep. I copied and pasted that from your code in the original post. That should be document.getElementsByTagName("body")[0].

It's getElementsByTagName (plural)
rgranlundAuthor Commented:
Thanks.  I did google that and fix it.  However, when I click on my link in my html it still goes to the txt page instead of staying on the index.html page.
Russ SuterSenior Software DeveloperCommented:
I assume you're clicking on the <a> element. It will follow the href unless you add a "return false;" in the onclick event handler like this:

<a href="file/ajax.txt" onclick="return false;">Load AJAX File</a>

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You may want to add a call to the javascript method containing the AJAX call before the return false.
Russ SuterSenior Software DeveloperCommented:
also another minor goof on my part. Since the dataType specified is "text" you need to modify the ajaxCallback method a little.

var pText = document.createTextNode(result.responseText);

should read

var pText = document.createTextNode(result);
Russ SuterSenior Software DeveloperCommented:
This example in its entirety works:

<!DOCTYPE html>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
    <title></title>
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
    <script type="text/javascript">
        function makeCall()
        {
            $.ajax({
                url: "AjaxTextFile.txt",
                dataType: "text",
                success: ajaxCallback
            });
        }

        function ajaxCallback(result, status, xhr)
        {
            var body = document.getElementsByTagName("body")[0];
            var p = document.createElement("p");
            var pText = document.createTextNode(result);
            p.appendChild(pText);
            body.appendChild(p);
        }
    </script>
</head>
<body>
    <a href="file/ajax.txt" onclick="makeCall(); return false;">Load AJAX File</a>
    <br /><br />

    <a href="file/ajax.txt">Load AJAX HTML</a>
</body>
</html>

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You'll need to modify your "url" parameter to point to the correct file.
rgranlundAuthor Commented:
Awesome!  Thanks.
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