Trouble Using a PHP Script to Send an Image

[Michael J. Astrauskas]

I have a script, which I've called test-loadpic.php and some pages
reference it by means of

<img src="test-loadpic.php?sou rcepic=$picNum" >

where $picNum stores a number. This part itself works fine.
test-loadpic.php uses the "sourcepic" GET variable to reference a file
on disk, get it's file/MIME type, transmit the headers and then the raw
image. The important code is below:

header ("Content-type: $fileMIME");
header ("Content-Length: ".filesize($fil ePath.$fileName ));

$fp = fopen ($filePath.$fil eName, "rb");
$er = fpassthru ($fp);
fclose($fp);

$fileMIME is the MIME type of the image, of course ("image/gif" or
"image/jpeg" in my case). $filePath and $fileName point to the image
(ex: "/var/www/phptest/images/" and "filter.jpg ").

Is this all I need to properly transmit an image?

This seems to work fine in IE but sometimes Netscape 7.1 tries to
download the image instead of showing it. Are my headers incorrect? I
suspect there's a glitch in my scripting that IE can tolerate but
Netscape can't. Is there a proper "end of image" code I need to send to
the browser that isn't included at the end of the images themselves?

Thank you in advance to anyone with some insight.

--
- Michael J. Astrauskas

[Shane Lahey]

yes should work fine...

you could try this function to display the image....
Usage: showImage('/path/to/image.jpg');
<?php

function showimage($path )
{
if (!file_exists($ path))
return false;

$filesize = filesize($path) ;
$basename = basename($path) ;

$info = @getimagesize($ path);
if (!is_array($inf o))
return false; // invalid image

switch ($info[2])
{
case 1:
$contentType = 'image/gif';
break;
case 2:
$contentType = 'image/jpeg';
break;
case 3:
$contentType = 'image/png';
break;
default:
return false; // unsupported
break;
}

$fd = @fopen($path, 'rb');
if (!is_resource($ fd))
return false; // could not read file.

$contents = '';
while ($data = fread($fd, 4096))
{
if (strlen($data) == 0)
break; // EOF
$contents .= $data;
}
fclose($fd);

Header('Content-type: ' . $contentType);
Header('Content-length: ' . $filesize);
Header('Content-Disposition: inline; filename='.$bas ename);
echo $contents;
return true;
}

?>
On Tue, 25 May 2004 10:42:31 -0700, "Michael J. Astrauskas"
<tr****@nospam. cox.net> wrote:

I have a script, which I've called test-loadpic.php and some pages
reference it by means of

<img src="test-loadpic.php?sou rcepic=$picNum" >

where $picNum stores a number. This part itself works fine.
test-loadpic.php uses the "sourcepic" GET variable to reference a file
on disk, get it's file/MIME type, transmit the headers and then the raw
image. The important code is below:

header ("Content-type: $fileMIME");
header ("Content-Length: ".filesize($fil ePath.$fileName ));

$fp = fopen ($filePath.$fil eName, "rb");
$er = fpassthru ($fp);
fclose($fp);

$fileMIME is the MIME type of the image, of course ("image/gif" or
"image/jpeg" in my case). $filePath and $fileName point to the image
(ex: "/var/www/phptest/images/" and "filter.jpg ").

Is this all I need to properly transmit an image?

This seems to work fine in IE but sometimes Netscape 7.1 tries to
download the image instead of showing it. Are my headers incorrect? I
suspect there's a glitch in my scripting that IE can tolerate but
Netscape can't. Is there a proper "end of image" code I need to send to
the browser that isn't included at the end of the images themselves?

Thank you in advance to anyone with some insight.

[Tim Van Wassenhove]

In article <rl************ *************** *****@4ax.com>, Shane Lahey wrote:

<?php

function showimage($path )
{
if (!file_exists($ path))
return false;

Assuming the $path is in public space

header('Locatio n: http://url/to/image');

Done ;)

--
Tim Van Wassenhove <http://home.mysth.be/~timvw/contact.php>

[Michael J. Astrauskas]

Shane Lahey wrote:

$contents = '';
while ($data = fread($fd, 4096))
{
if (strlen($data) == 0)
break; // EOF
$contents .= $data;
}
fclose($fd);

What is the purpose of this loop?

--
- Michael J. Astrauskas

[Shane Lahey]

On Tue, 25 May 2004 22:48:42 -0700, "Michael J. Astrauskas"
<tr****@nospam. cox.net> wrote:

Shane Lahey wrote:

$contents = '';
while ($data = fread($fd, 4096))
{
if (strlen($data) == 0)
break; // EOF
$contents .= $data;
}
fclose($fd);

What is the purpose of this loop?

binary safe way to read the contents of a file into a variable.......

[Dennis Biletsky]

"Shane Lahey" <s.*****@roadru nner.nf.net>
news:rl******** *************** *********@4ax.c om...

yes should work fine...
[skip] $fd = @fopen($path, 'rb');
if (!is_resource($ fd))
return false; // could not read file.

$contents = '';
while ($data = fread($fd, 4096))
{
if (strlen($data) == 0)
break; // EOF
$contents .= $data;
}
fclose($fd);

Header('Content-type: ' . $contentType);
Header('Content-length: ' . $filesize);
Header('Content-Disposition: inline; filename='.$bas ename);
echo $contents;
return true;
}

?>

Header('Content-type: ' . $contentType);
readfile($path) ;
I'm using a little bit shorter script. Would you like to compare with yours?
I'm interesting in your opinion and others opinion too.

[Shane Lahey]

heh, sorry I don't actually have one, I came up with this off the top
of my head :/

On Wed, 26 May 2004 10:19:46 +0300, "Dennis Biletsky" <uf***@ua.fm>
wrote:

"Shane Lahey" <s.*****@roadru nner.nf.net>
news:rl******* *************** **********@4ax. com...

yes should work fine...

[skip]

$fd = @fopen($path, 'rb');
if (!is_resource($ fd))
return false; // could not read file.

$contents = '';
while ($data = fread($fd, 4096))
{
if (strlen($data) == 0)
break; // EOF
$contents .= $data;
}
fclose($fd);

Header('Content-type: ' . $contentType);
Header('Content-length: ' . $filesize);
Header('Content-Disposition: inline; filename='.$bas ename);
echo $contents;
return true;
}

?>

Header('Conten t-type: ' . $contentType);
readfile($path );
I'm using a little bit shorter script. Would you like to compare with yours?
I'm interesting in your opinion and others opinion too.

[Michael J. Astrauskas]

Tim Van Wassenhove wrote:

Shane Lahey wrote:

<?php

function showimage($path )
{
if (!file_exists($ path))
return false;

Assuming the $path is in public space

header('Locatio n: http://url/to/image');

Done ;)

This is a good idea, and I'll keep it in mind, but I'm trying to hide
file names from users.

--
- Michael J. Astrauskas