Hello,
I guess anyone have made the function to coppy directories, not files.
I'd like to copy directory including sub directories to other place.
Doesn't anyone has mades that kind of function?And please telll me.
Thamks in advance.
HiroshiAyukawa
http://hoover.ktplan.ne.jp/kaihatsu/php_e
I have a new very intriguing problem at hand.
I have the name of my Images stored in my mySQL
database in one column. Now when i pick the images,
they are displayed as it as. However, they are the big
images, and the thumbnails of those images are stored
with an "a" at the end of thier names.
T
(untested)
$newname = eregi_replace('\.jpg$', 'a.jpg', $oldname);
No point messing up your database; just use something like the above when
you're outputting the image tags for the thumbnails.
miguel
On Mon, 8 Apr 2002, Thomas Edison Jr. wrote:
> I have a new very intriguing problem at han
$parts = preg_split('/\./', $filename);
$new_filename = $parts[0] . "a.jpg";
(Not tested, but _should_ work.)
/bsh/
Thomas Edison Jr. wrote:
> I have a new very intriguing problem at hand.
>
> I have the name of my Images stored in my mySQL
> database in one column. Now when i pick the images
On Monday 08 April 2002 15:41, Thomas Edison Jr. wrote:
> How can i :
> 1. Insert "a" at the end of the name of the image,
> before the ".extension" through PHP.
> The problems are that the names are stored in the
> database WITH the extesion. And the extensions also
> vary, some are JPG and some
#
#
# boolean copy_dirs ( string src_dir, string target_dir )
#
#
# copy shopdirectories into a new shop
#
# Function Parameters
there was a typo...this one should work
#
#
# boolean copy_dirs ( string src_dir, string target_dir )
#
#
# copy a directory with subdirectories to a target directory
Hi,
Thanks for your relies. There are a couple of
problems. Firstly, i'm picking the filename from the
database, so my code is something like this :
if ($myrow = mysql_fetch_array($result)) {
do {
echo("
wrote:
> (untested)
>
> $newname = eregi_replace('\.jpg$', 'a.jpg',
> $oldname);
>
THANKS!
That works!!
Except on PHP side I have to produce javascript code without tag,
like
xTest.php - SERVER1 with PHP
document.write();
on server 2 I just have to write this inside
