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How to load php dynamic pages using AJax Tab

Hi, I have been having problems loading a php page using the ajax tab option. Below is an example webpage and here is the tab anchor I am using:


<a href="includes/your-account-includes/uploaded-files.php" data-ajax="{target: 'view3', id: 'fragment1'}">Previously Uploaded Files</a>

<div id="view3"> </div>
Will  10 years ago   viewed: 19988    

5 Answers

Your page /includes/your-account-includes/uploaded-files.php contains only the <h2 align=center>Uploaded Files</h2> and it is retrieved and displayed. Nothing wrong.
Milo   10 years ago
In actuality this is the uploaded-files.php. What's being displayed is the html mark up and not the php.

<h2 align=center>Uploaded Files</h2>
//This is for displaying uploaded templates
include_once ($_SERVER['DOCUMENT_ROOT']."/includes/db_connect.php");
// Check connection
if (mysqli_connect_errno())
 echo "Failed to connect to MySQL: " . mysqli_connect_error();
  $result = mysqli_query($con,"SELECT * FROM images WHERE UserName='$Username'");
while($row = mysqli_fetch_array($result)){

<a href="<?php echo $row['ImageLink']?>" onclick="NewWindow(this.href,'mywin','500','500','no','center');return false" onfocus="this.blur()">
$address = $row['ImageLink'];
$last3chars = substr($address, -3); 

    echo "<img src=$row[ImageLink] width=80> </a>  <a href=zip.php?Option1=$row[ImageLink]>[Use This Template]</a> ";


    //the path to the PDF file
    $strPDF = $row['ImageLink'];
    $imagelinkname =(rand(10000000,999999999));
    exec("convert \"{$strPDF}[0]\" -colorspace RGB -geometry 200 \"_optionfiles/step1/$imagelinkname.gif\"");
    echo "<img src='_optionfiles/step1/$imagelinkname.gif' width='80'> </a> <a href=zip.php?Option1=$row[ImageLink]>[Use This Template]</a> ";
Will   10 years ago
No matter what your PHP code is, the page will render out the HTML code, and the rendered HTML code should contain:
<div id="fragment1">....(the fragment content)....</div>. From your page I only find <h2 align=center>Uploaded Files</h2>. So you need to double check your PHP code which I am not good at.
Milo   10 years ago
Okay got it to work lol. It was indeed the session. I guess when the page is called the current session isn't carried over... So for future reference, if anybody has php that requires session data... my solution was to unset and set the session:
session_start();$Username = $_SESSION['Username'];
Will   10 years ago
After a few tests- the issue is with the session. The Username is not being recognized- though, it is assigned correctly.
Will   10 years ago


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