Hello,
I'm creating a new website for showing my photo's on a webpage. The first code for getting the photo's (jpg files) out off a directory and show it on a webpage works (on the provider I don't have a MySQL database, so I need to work with directories).
I also created a link to the image for showing it in a new page. But now I want to use another page with some text, border etc ... for showing the image.
After searching on the net I found that you can do it with passing arguments like. "weergeven.php?ARG=fungi.jpg" and then write in the weergeven.php "echo $ARG".
But it doesn't work, there is no value in the variable $ARG.
Can someone help me with this problem or is there a more elegant solution ?
Thanks,
Christof Laeremans
Showing images
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ChristofLaeremans
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malcolmboston
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Code: Select all
if (isset($_GET['ARG']))
{
print "<img src=\"{$_GET['ARG']}\">";
}
else
{
print "argument not passed";
}-
ChristofLaeremans
- Forum Newbie
- Posts: 2
- Joined: Fri May 06, 2005 1:25 am
- Location: Tienen, Belgium