error help - Dynamic Image
Posted on
16th Feb 2014 07:03 pm by
admin
I've been working on making my site less cluttered in the directories and more secure lately. In an attempt to prevent bandwidth theft from other people using images, I decided to switch all my site images to a dynamic image file. I've written a small test code to iron out as many bugs as I can BEFORE I impliment it fully.
Code:
Code: <?php
if ($_SESSION['i'] != "yes"}
{
$im = "theft";
}
else
{
$im = $_GET['i'];
}
switch($im)
{
case 'theft":
$image = imageCreateFromJPG('/theft.jpg');
break;
case 'logo':
$image = imageCreateFromJPG('/logo.jpg');
break;
}
imageJPG($image);
imageDestroy($image);
?>
When using it in the test copy of my site's index page (calling for the logo image) it works fine.
When I try opening the page (with the GET['i'] set to logo, theft, and not set at all), I get the following error:
QuoteParse error: syntax error, unexpected T_IF in /home/ck9/public_html/media/images/site.php on line 3
I tried setting the current If...else as a long comment and re-writting it another way, but that didn't fix the issue. I tried looking up the error in the PHP manual, but the information presented wasn't helpful to me.
Thanks in advance for any help you guys can provide (and the help here has always been great
No comments posted yet
Your Answer:
Login to answer
58
42
Other forums
Slashes
Have a small problem and I'm not able to understand why I'm getting the results I'm getting... and i
Preloading images
Posting this question here because I am not sure where this should belong.I am building an asp.net a
Code clarification
Hi
In the following code what could be the "search_print()" and where it could be
form class help (oop php5)
Hidy Ho Neighbors,
I'm forcing myself to learn oop/classes for php5. It seems like a good id
Help with email validation please...
Hi,
Please could you help.
I have a register.php login page where users register, the detail
Request-URI Too Large
I have created a simple submit form for a mysql database that puts a piece of code into database.<
serializing objects - loses methods - the point being?
Hi
I know that serializing an object will lose the methods. But whats the point in that? What
$action = "insert"; //$action = $_GET['action'];
$action = "insert";
//$action = $_GET['action'];
why is this invalid type? I am
CODE NOT WORKING
Code: [Select]<?php
//include shared codes
include '../lib/common.php';
include
mysql query with single quotes in a variable
$sitedetails = "INSERT INTO vars (address, sitename, description, ownername, theme) VALUES ('$u