View Full Version : Can someone tell me how this is not displaying any output?

10-11-2004, 03:49 AM
I have a website that I have design that is suppose to pull information out of a database. Well when someone clicks on a link it is suppose to change the content of the website. Inside of the database in a table called Land_Pictures is this...
<form action="home.com/index.php" method="get" name="picture1">
<? $img=1.jpg ?>
<a href="" onClick="window.open('image.php?id=<?echo $img;?>', 'win1', 'width=300,height=350')">
<img src="<?echo $img;?>.jpg" height="221" width="200" border="0">

Well this is the same code that I have to work with other pictures. I also have other buttons that display other information from the database, so I know it is not my connection string or anything, but what this does is just show sort of an image place holder, but any help would be greatly appreciated...thanks


10-11-2004, 04:36 AM
<? $img=1.jpg ?>
should be
<? $img="1.jpg"; ?>

And I am guessing this is wrong unless your image is called '1.jpg.jpg'
<img src="<?echo $img;?>.jpg" height="221" width="200" border="0">

you are also missing a trailing </a> and I am unsure what the form is doing there ?

10-11-2004, 04:46 AM
I ended up taking the form out, but what the form did was return it back to the main window...this works sort of like a pop up but I did was take the $img="1.jpg? and make it equal just 1
Now the image place holder is still there, but when I go and look at the properties on that image it gives me this as the location, if that helps any...
now my code reads this
<? $img="1"; ?><a href="" onClick="window.open('image.php?id=<?echo $img;?>', 'win1', 'width=300,height=350')"><img src="<?echo $img;?>.jpg" height="221" width="200" border="0"></a>

10-11-2004, 05:26 AM
well I just checked your site and it appears to be working now ? /index.php at least