Question

Am fairly new to PHP and am making a basic CRUD style management system. I Have an update page and it displays data from a News table, and populates a form with it. The current picture ?(reference) is pulled through and displayed on the form. However if a user wants to change the picture they can press a ‘delete’ button and then I have written some PHP to display a upload button, set the values in the database for the image to null and hide the delete button, allowing the user to upload a new picture.

The Delete button only removes the reference (path) to the picture from the database, it doesn’t delete the actual picture.

This is the HTML control to show the image and delete button. It also shows how the delete button works:

 <td align="right">Image 1:</td>
 <td align="left"><img src="uploads/newsimages/<?php echo $row["Image"]; ?>" width="230" border="0">&nbsp;<a href="UpdateNews.php?change=imagex&cid=<?php echo $row["NewsID"]; ?>">delete</a></td>

As you can see, when clicked it sets change=imagex and cid= the current news id.

There is then an if statement I have written, but it doesn’t seem to only get activated when the delete button is clicked. Because I always get an error that ‘cid’ is undefined. It is as follows:

<?php
if (isset($_GET['change'] = "image1") {
    $query = "UPDATE  Table_Name SET Image = '' WHERE NewsID =".$_GET['cid']." ";
}

?>

I am pretty sure my lack of PHP knowledge is letting me down and I am trying to go about this the wrong way, because however I alter the if statement it always gives me an error. First it was cid is undefined so I changed to id but i already use that for something else, another query/function. I hope that all amde sense, can anyone tell me where Im going wrong?

Answer 1

You are missing a parenthesis + you have to specify individually:

if (isset($_GET['change'] = "image1") {

Change to:

if (isset($_GET['change']) && $_GET['change'] == "image1") {