On 5/18/06, Gustav Wiberg <[EMAIL PROTECTED]> wrote:
Hi!
The thing I want to do is to copy a picturefile to another picturefile.
The thing is that I want to copy this file, show it , and then delete it
(when it has been shown) .Is this possible?
I want to do this, because of avoiding problems with cache when uploading
file through an admin-online-system... (the customer uses IE)
When I delete the file in code down below, the picture is not shown (I guess
because the browser hasn't rendered out all info?)
If you want more code, tell me :-)
Best regards
/Gustav Wiberg
$fileName = "pictures/products/$dbIDProduct1" . "_small";
$ran = strval(mktime()); //Current time
if (file_exists($fileName . ".gif")) {
copy($fileName . ".gif", "pictures/products/1_$ran.gif");
showpicture("pictures/products/1_$ran.gif", $dbProductName1, 300, 150,
"top");
//deletefile("pictures/products/1_$ran.gif");
}
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I had to do something similar to keep a javascript file from being
cached by an aol proxy. Heres a simple example that you could change
to suit your needs by changing the js to something like image.php and
making it send image headers and outputting the contents of your file.
1) Rewrite rule in httpd.conf or virtual.conf
=======================================================================
RewriteRule ^/lib/(.*)/test.js.php /lib/test.js.php [QSA,L]
2) Create test.js.php in /lib/test.js.php
=======================================================================
<?php
header('Content-Type: text/javascript');
?>
3) Access url: http://example.local/lib/4908574987/test.js.php. You
can create your url like /lib/time()/test.js.php.
This way you don't have to worry about creating and deleting files.
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