Displaying Images form a Directory

[Jameson]

Happy New Year, Everyone!

I am trying to figure out how to display a bunch of images (mainly
JPEGs, but possibly a few GIFs and PNGs as well) that are stored in a
local directory on the system. I can do this with the glob() function,
but I can't seem to put in a directory other than one within the
webroot. For example, I can only put "/uploads" and not
"/Volumes/jray/Pictures...".

Any ideas how to get around this? If I can't use the glob function,
it's fine, but I only want images to be displayed (and not any other
file that happpens to be stored in that directory).

Thanks in advance!

[witkey]

<?php
// An Example about listing Images.
$dir = ".";
$odir = opendir($dir);
while($file = readdir($odir)) {
if(filetype($fi le) == "image/JPEG") {
echo $file;
}
?>

[Jameson]

Thanks for the fast response. What I'm trying to do is actually display
the images, not list them. What I've got so far (sorry for not posting
this earlier) is:

<?php
echo "<table><tr >";

foreach(glob("u ploads/*.{jpg,JPG,jpeg ,JPEG,gif,GIF,p ng,PNG}",
GLOB_BRACE) as $images)
{
echo "<td><img src=\"".$images ."\"><br/>";
}

echo "</tr></table>";
?>

It works just fine if I want to view images in the "uploads/" folder
(which is parallel to this script), but if I want to change it to be
"/Volumes/jray/Pictures" it won't work.

witkey wrote:

<?php
// An Example about listing Images.
$dir = ".";
$odir = opendir($dir);
while($file = readdir($odir)) {
if(filetype($fi le) == "image/JPEG") {
echo $file;
}
?>

[Chung Leong]

Well, if the images aren't within the webroot, then the web server
won't serve them...

[Jim Carlock]

"Jameson" <ja*********@co mcast.net> wrote:

Thanks for the fast response. What I'm trying to do is
actually display the images, not list them. What I've
got so far (sorry for not posting this earlier) is:

Use witkey's suggestion with a modification?

<?php
// An Example about listing Images.
$dir = ".";
$odir = opendir($dir);
while($file = readdir($odir)) {
if(filetype($fi le) == "image/JPEG") {
$sAltText = '"Whatever the alt text is if so required."';
echo '<img src="' . $dir . $file . ' border="0" alt="' . $sAltText . '" /><br />' . "\n";
}
?>

Jim Carlock
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[Jerry Stuckle]

Chung Leong wrote:

Well, if the images aren't within the webroot, then the web server
won't serve them...

Not true. You can serve them through PHP, e.g.

<img src="showimg.ph p"...>

And showimg.php can be something like:

header('Content-type: image/gif');
header('Content-length: '.filesize($img _filename));
$file_pointer = fopen('/some/other/directory/img.gif', 'rb');
fpassthru($file _pointer);
fclose($file_po inter);

The graphic can reside anywhere on the system which is accessible to the
Apache process (not just under DocumentRoot).

--
=============== ===
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
js*******@attgl obal.net
=============== ===

[Jameson]

Hi Jim:

Thanks for the suggestion. I tried it out, but still got a blank
screen. I think Chung might be right, because if I try "<img
src="/Library/WebServer/Documents/uploads/door.jpg>" the web server
won't serve them. I think the client's browser is trying to find the
images locally on the HD.

Maybe what I'm trying to do isn't possible. Do you think using the
glob() function within the opendir() function might work?

Thanks for all of the help!

Jim Carlock wrote:

"Jameson" <ja*********@co mcast.net> wrote:

Thanks for the fast response. What I'm trying to do is
actually display the images, not list them. What I've
got so far (sorry for not posting this earlier) is:

Use witkey's suggestion with a modification?

<?php
// An Example about listing Images.
$dir = ".";
$odir = opendir($dir);
while($file = readdir($odir)) {
if(filetype($fi le) == "image/JPEG") {
$sAltText = '"Whatever the alt text is if so required."';
echo '<img src="' . $dir . $file . ' border="0" alt="' . $sAltText . '" /><br />' . "\n";
}
?>

Jim Carlock
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[Jim Carlock]

Hi Jameson,

<g> I feel like I'm talking to a son... (j.k)

I didn't test the witkey's code. It appears that filetype()
returns "file" or "dir". Try the following.

<?php
// An Example about listing Images.
$dir = ".";
$dh = opendir($dir);
// skip the . and ..
$file = readdir($dh);
$sAltText = 'Picture';
// skip . and ..
if ($file == ".") {
$file = readdir($dh); $file = readdir($dh);
do {
if (mime_content_t ype($file)=="im age/jpeg") {
echo '<img src="' . $dir . $file . '" border="0" alt="' . $sAltText . '" /><br />' . "\n";
} while($file = readdir($dh));
}
}
closedir($dh);
?>

That should work I think. It's air code. I tested it out here
but I'm running into a problem where the mime_content_ty pe
is a problematic function right at the moment.

I read that the glob function is supposed to be a better
alternative in the PHP manual. And if you know that
all the files in the folder are jpg's then you won't have to
do the mime_content_ty pe test.

My apologies about not testing witkey's code. Hope this helps.

If anyone knows how to get the php_mime_magic. dll
file to work on a Windows system, I could use the help.
I uncommented the line in php.ini...

extension=php_m ime_magic.dll

and put the php_mime_magic. dll in the system32 folder.
Then I stopped apache and restarted apache...

net stop apache
net start apache

Thanks, much.

Jim Carlock
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[Jameson]

Hi Jim:

Thanks for the new code. I think we're getting closer, but it still
doesn't seem to be working. I added the error reporting code below to
the file, but it doesn't tell me that anything is wrong. It's strange,
because I don't even see an icon from the browser saying that it can't
find the image. Do you think I could be missing some component of my
PHP installation? I know the directory path is fine, because I can get
to it right from the terminal on the computer.

ini_set("displa y_errors",true) ;
error_reporting (E_ALL );

Any more ideas? Do you think if I used the glob() function within the
opendir() function, it might work?

Thanks again for the help. I really appreciate it!

[Jim Carlock]

"Jameson" <ja*********@co mcast.net> wrote:

Thanks for the new code. I think we're getting closer, but it still
doesn't seem to be working.
Hi Ray,

Are you working in the MacIntosh environment? I'm working on
with a PC. On the PC, the php.ini file typically gets found in the
system32 folder. I'm not sure how things work on a MacIntosh,
so perhaps you can explain the php.ini concepts relating to
MacIntosh (if that is the case) for my benefit. The php stuff is
very new to me.

I did get the following code to work here, using glob(). I really
like this better than having to do mime_content_ty pe stuff. And
so much in the PC and unix world seem very much related,
using file extensions to denote the contents of the file.

<html>
<head>
<title>List of Files</title>
</head>
<body>
<p><?php
$dir = '.';
$sAltText = "Picture";
foreach (glob("*.jpg") as $file) {
echo "file: $file<br />\n";
echo "<i>filenam e:</i> <b>$file</b>, <i>filetype:</i> <b>" . filetype($file) . "</b><br />\n";
echo '<img src="' . $file . '" border="0" alt="$sAltText" /><br />' . "\n";
}
?></p>
</body></html>
I added the error reporting code below to the file, but it doesn't
tell me that anything is wrong. It's strange, because I don't even
see an icon from the browser saying that it can't find the image.
When you view the page through the browser, take a look at the
source code (the HTML output). The code I posted previously
seems to have had a small bug in one of the lines, and I gave up
testing it when I couldn't get the php_mime_magic. dll to provide
the php_mime_conten t() function. So I don't know how the
Mac environment works, nor the Unix environment. The file name
could possibly be of a different extension in each environment. If
anyone else here can provide a little help here that would be
great.
Do you think I could be missing some component of my PHP
installation? I know the directory path is fine, because I can get
to it right from the terminal on the computer.
ini_set("displa y_errors",true) ;
error_reporting (E_ALL );
Any more ideas? Do you think if I used the glob() function within
the opendir() function, it might work?
glob() is alot easier... there's no need for opendir() and readdir()
when using the glob function.

I just tested the following out and it works quite well.

<?php
$dir = '.';
$sAltText = "Picture";
foreach (glob("*.jpg") as $file) {
echo "file: $file<br />\n";
echo "<i>filenam e:</i> <b>$file</b>, <i>filetype:</i> <b>" . filetype($file) . "</b><br />\n";
echo '<img src="' . $file . '" border="0" alt="$sAltText" /><br />' . "\n";
}
?>
Thanks again for the help. I really appreciate it!

You're welcome. I'm learning as well. So thanks back at you!

My mime_content_ty pe() function is failing I think, because
the directory path for the php install folder is set to C:\php4
and php isn't installed there. So I guess I need to find the
source code for that particular DLL and modify the source
code. If anyone knows where to pick up the source code
for the php_mime_magic. dll feel free to leave a hint. It's a
shame (for me and others) that the people that compiled it
used an absolute path inside the DLL file.

Jim Carlock
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