stackoverflow. I encountered this error, when i try to access and retrieved the data back from the database. to show in the table. but, after i run i get this error. i really hope that i can a solution.
here is my service.php code
<?php
//sambung ke pangkalan data
require('config.php');
include('session.php');
$photographerID = $_SESSION['photographerID'];
$serviceID = $_SESSION['serviceID'];
//memastikan pengguna login terlebih dahulu
?>
<!--HTML Bermula-->
<html>
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<head>
<title>DETAILED SERVICE</title>
<br>
<center><h2 style="color: #00004d;">DETAILED SERVICE</h2>
<fieldset>
<table width="auto" border="1" class="imagetable">
<?php
$dataA=mysqli_query($conn,"select * from serviceadvertizing where serviceID='$serviceID'");
$infoA=mysqli_fetch_array($dataA);
?>
<tr>
<td width="200" style="background-color: #FF8500; color: white; text-align: center; height: 50px;" ><b>Service ID </b></td>
<td style="width: auto; padding-left: 5px; height: auto;"><?php echo $infoA['serviceID'];?></td>
</tr>
<tr>
<td width="200" style="background-color: #FF8500; color: white; text-align: center; height: 50px;" ><b>Photographer ID </b></td>
<td style="width: auto; padding-left: 5px; height: auto;"><?php echo $infoA['photographerID']; ?></td>
</tr>
</table>
</fieldset>
</body>
</html>
here is my session.php code:
session_start();
if(!isset($_SESSION["photographerID"]) && !isset($_SESSION["artworkID"]) && !isset($_SESSION["serviceID"]) && !isset($_SESSION["Course_ID"])){
echo "kjhfdfm,";
header("Location:index.php");
//exit();
}
i already tried many solution, but i did not found any. the service id is exist in the database but why still error ? or i have a problem in the php query ?