Displaying an HTML popup using AJAX

Question

I'm trying to make an HTML page that displays another html file in an alert; but it's not displaying when the triggering button is pressed.

<html>
    <head>
        <script>

        var xmlhttp=new XMLHttpRequest();

        function pop() {
            xmlhttp.open("GET","content.html",true);
            xmlhttp.send();
            xmlhttp.onreadystatechange=function() {
                if(xmlhttp.readystate==4&&xmlhttp.status==200) {
                    alert(xmlhttp.responseText);
                }
            }
        }

        </script>
    </head>
    <body>
        <input type="button" name="test" value="push" onclick="pop()">
    </body>
</html>

Here is the content of content.html

<html>
    <body>
        Hi!
    </body>
</html>

You might want to use jQuery. Saves you some work and your code will be much nicer to read. api.jquery.com/category/ajax — ThiefMaster
– ThiefMaster, Commented Jul 18, 2011 at 9:32
How can i obtian Hi! only instead of the whole html code???thanks — Leo Chan
– Leo Chan, Commented Jul 18, 2011 at 9:39
Yup i would like to learning ajax and i am finding some ajax exercise online. — Leo Chan
– Leo Chan, Commented Jul 18, 2011 at 9:40
We appreciate it when homework is actually tagged as homework. =) — J. Steen
– J. Steen, Commented Jul 18, 2011 at 10:09

pimvdb · Accepted Answer · 2011-07-18 09:33:01Z

1

In fact it's readyState. JavaScript is case-sensitive.

Also, it might be better to send after setting up everything.

Lastly, you have a missing }.

var xmlhttp=new XMLHttpRequest();

function pop()
{
 xmlhttp.open("GET","content.html",true);
 xmlhttp.onreadystatechange=function()
 { if(xmlhttp.readyState==4&&xmlhttp.status==200)
    {alert(xmlhttp.responseText);}
 }
 xmlhttp.send();
}

answered Jul 18, 2011 at 9:33

pimvdb

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1 Comment

Leo Chan Over a year ago

alright. Thanks everyone. It seems I made some careless mistake.

Tatu Ulmanen · Accepted Answer · 2011-07-18 09:32:00Z

0

Yep, i'm counting three opening braces ({) but only two closing braces (}). Check your browser's error console to spot such errors.

answered Jul 18, 2011 at 9:32

Tatu Ulmanen

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Comments

Saurabh · Accepted Answer · 2011-07-18 09:32:57Z

0

Check Below: Closing bracket is missing.

var xmlhttp=new XMLHttpRequest();

function pop()
{
   xmlhttp.open("GET","content.html",true);
  xmlhttp.send();
  xmlhttp.onreadystatechange=function()
  {

if(xmlhttp.readystate==4&&xmlhttp.status==200)
   {
    alert(xmlhttp.responseText);
   }
  }
}

answered Jul 18, 2011 at 9:32

Saurabh

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