php / Javascript / Ajax loading content does not work

Question

I'm trying to load content from a php file which name is "include.php" to a in another php file which name is "index.php". But the loading does not work. The code is as below:

The file: index.php

<header>
<script type="text/javascript">
    function load(){
        if (window.XMLHttpRequest) {
            xmlhttp = new XMLHttpRequest();
        }else{
            xmlhttp = new ActiveXObject('Microsoft.XMLHTTP');
        }

        xmlhttp.onreadystatechage = function (){
            if(xmlhttp.readyState == 4 && xmlhttp.status == 200){
                document.getElementById(adiv).innerHTML = xmlhttp.responseText;
            }
       }

        xmlhttp.open('GET', 'include.php', true);
        xmlhttp.send();
    }

    </script>
</head>
<body>
    <input type="submit" value="Submit" onclick="load();">
    <div id="adiv"></div>
</body>

The File: include.php

<?php
    echo 'Hello!';
?>

Thanks.

Go to Inspect element -> network tab -> what can you see there? — Kunal Pradhan
– Kunal Pradhan, Commented Jun 21, 2015 at 6:59
There is a list of my three files: "index.php", "style.css", and"include.inc.php". Method of all these three files is "GET". STATUS for "index.php" and "include.inc.php" is "OK". STATUS of "style.css" is "Not Modified." — Sara
– Sara, Commented Jun 21, 2015 at 20:35

Michael Christensen · Accepted Answer · 2015-06-21 07:13:40Z

1

If what you are doing is the way you describe then this is simpler:

<header>
<?php include("include.php") ?>
</head>
<body>
    <input type="submit" value="Submit" onclick="load();">
    <div id="adiv"></div>
</body>

If you want to get result from include.php into JavaScript then you'll probably be better using ajax.

By the way, if you are planning a "universal header" for all your PHP files, you don't need to echo it just write it as normal HTML with any necessary PHP tags

edited Jun 21, 2015 at 7:13

answered Jun 21, 2015 at 7:08

Michael Christensen

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2 Comments

Michael Christensen Over a year ago

Hey if you like the answer can you give a check mark to signify this question has been answered?

Sara Over a year ago

Thanks Michael. What I want to do is loading some content from another file to a div without refresh the page. In this case, when I click the "Submit" button, the content on another file should be loaded under the "Submit" button but no page refresh.

Nitesh Kumar Anand · Accepted Answer · 2015-06-21 07:04:57Z

0

Should it not be document.getElementById("adiv").innerHTML = xmlhttp.responseText;? Notice the quotes.

answered Jun 21, 2015 at 7:04

Nitesh Kumar Anand

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2 Comments

Sara Over a year ago

Thanks, and I correct this error, but still does not work.

Sara Over a year ago

It works now. I changed xmlhttp.onReadyStateChange = function (){ back to xmlhttp.onreadystatechange & corrected getElementById("adiv"). Thanks all of your helps. @Nitesh Kumar Anand, @Kirs Sudh, @Michael Christensen, @ Kunal Pradhan

Kirs Sudh · Accepted Answer · 2015-06-21 07:12:57Z

0

....xmlhttp.onreadystatechage = function (){ if(xmlhttp.readyState == 4 && xmlhttp.st...

Check the spelling onReadyStateChange.

answered Jun 21, 2015 at 7:12

Kirs Sudh

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