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Hi All,

I have problem with my dynamic page, in the index.php i have 3 link that connect to database and each one has ID , when I click on hyperlink the dynamic page show on the other page on URL with ID like this : page01.php?id=6 , and change in to page01.php, the problem is that : how can get this ID in the page01.php to show only that specific ID information such as Image and description. it's like news page, when click on news link , next page just show description of that link.

thank you

$p = $_GET['page'];
		$page = "Applications/XAMPP/xamppfiles/htdocs/register/".$p.".php";
			echo "This is Home Page";
<a href="page01.php?id=<?php echo $row['id']?>">



if (isset($_GET['ID'])) {
    $id  = mysql_real_escape_string($_GET['ID']);
$result = mysqli_query($con,"SELECT * FROM upload WHERE id='1' ORDER BY id DESC");
		 if (mysql_num_rows() > 0) {
        $upload = mysql_fetch_object($res);
   echo "<h2>" . $rows['Image'] . "</h2>";

    else {
        // member does not exist with ID
Posted 19-Apr-13 23:11pm

1 solution

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Solution 2

this the way to solve this question check this out the video

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