Look at your code -- where do you tell it to set $file_name equal to the
file handles within e:\work\images? $file_name is only being set to the
values within e:\work -- not e:\work\images.
If you only need to display the images within e:\work\images, then why not
just set $dir_name to that in the first place? Otherwise (I believe) you're
going to have to use two separate while ($file_name=readdir($dir)) clauses
(or nest them, depending on what you really want to output)
hth
--kurt
----- Original Message -----
From: "McShen" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Saturday, July 07, 2001 8:38 AM
Subject: [PHP] chdir() help
hi
Currently, i am working in e:\work, and there is a folder named "image"
under e:\work, so, the path to image is e:\work\image.
I have a script under e:\work, and i wanna display all images under the
folder e:\work\image, I use chdir() to change the directory. but it wouldn't
work. it still displays the images under e:\work. Why is that? Does chdir()
work under win2k pro? here is my script
-------
<?php
$dir_name = "e:\work";
$dir = opendir($dir_name);
while ($file_name=readdir($dir)) {
if (($file_name!="." && $file_name!="..")) {
echo $file_name."\n ";
if (chdir('e:\work\image')) {
echo "current dir is e:\work\image";
}
echo "<IMG SRC=$file_name>";
}
}
closedir($dir);
?>
-----
Please help me to fix the problem. Thanks.
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