Re: [PHP] Re: Displaying images
Miller, Terion wrote:
>
>
> Well I have tried Numerous scripts and ways and still can't get the image to
> display.
>
> I have echoed the file and have been able to get the gibberish image code to
> display but not a real image, here is my full code if anyone wants to have a
> look, I am going crossed eyed here.
>
>
>
>
>
> These are just a few of the ways I have been trying to get the img tag to
> work on my output.php page:
>
>Scout Photo:
> echo$row['photoName']; ?> "\nScout Photo : "." "\">"; ?> src=\"image.php?filename=".$row['photoName']."\">\n"; ?>
> \n"; ?>
>
>
>
> Here is one way I was trying to get to work for the image.php page where the
> headers are supposed to work and don't
>
> include("inc/dbconn_open.php");
> error_reporting(E_ALL);
>
>
> //if (isset($_FILES['ePhoto'])){$ePhoto = $_FILES['ePhoto'];} else {$ePhoto
> ="";}
>
> $filename = $_GET['$filename'];
>
> $image = stripslashes($_REQUEST[photoName]);
>
> $sql = "SELECT ePhoto, photoName, photoType from eagleProjects WHERE
> photoName = $filename";
> $result=mysql_query($sql);
> $data = mysql_fetch_array ($result);
>
>
>
> $type = $data['photoType'];
> $name = $data['photoName'];
>
> header("Content-type: $type");
> header("Content-Disposition: attachment; filename=$name");
>
>
> echo $data["photoName"];
> echo $data["ePhoto"];
>
> exit;
It's hard to tell, not knowing what the db fields contain, but the two
glaring issues are:
1. unless photoType is image/something then it won't work. It can't
just be jpeg or gif. So maybe do header("Content-type: image/$type");
assuming $type is correct.
2. Since $data["photoName"]; is not of type image/whatever, the echo may
break the display.
--
Thanks!
-Shawn
http://www.spidean.com
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Re: [PHP] Re: Displaying images
On 5/28/09 2:09 PM, "Bastien Koert" wrote:
On Thu, May 28, 2009 at 3:06 PM, Miller, Terion
wrote:
Well I have tried Numerous scripts and ways and still can't get the image to
display.
I have echoed the file and have been able to get the gibberish image code to
display but not a real image, here is my full code if anyone wants to have a
look, I am going crossed eyed here.
These are just a few of the ways I have been trying to get the img tag to work
on my output.php page:
Scout Photo:Scout Photo : ".""; ?> \n"; ?>
\n"; ?>
Here is one way I was trying to get to work for the image.php page where the
headers are supposed to work and don't
include("inc/dbconn_open.php");
error_reporting(E_ALL);
//if (isset($_FILES['ePhoto'])){$ePhoto = $_FILES['ePhoto'];} else {$ePhoto
="";}
$filename = $_GET['$filename'];
$image = stripslashes($_REQUEST[photoName]);
$sql = "SELECT ePhoto, photoName, photoType from eagleProjects WHERE photoName
= $filename";
$result=mysql_query($sql);
$data = mysql_fetch_array ($result);
$type = $data['photoType'];
$name = $data['photoName'];
header("Content-type: $type");
header("Content-Disposition: attachment; filename=$name");
echo $data["photoName"];
echo $data["ePhoto"];
exit;
Terin,
I just gave the same code to another person on another forum and he had no
issues getting that to work. If it does not work, you will need to ensure that
the value you pass to the show_image file is the same value that you are using
to look up the image. So if its an id number make sure you are using the same
id field to look up the image.
Hi Bastien,
I am using the ID to call the image any other ideas...
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Re: [PHP] Re: Displaying images
On Thu, May 28, 2009 at 3:06 PM, Miller, Terion <
tmil...@springfi.gannett.com> wrote:
>
>
>
> Well I have tried Numerous scripts and ways and still can't get the image
> to display.
>
> I have echoed the file and have been able to get the gibberish image code
> to display but not a real image, here is my full code if anyone wants to
> have a look, I am going crossed eyed here.
>
>
>
>
>
> These are just a few of the ways I have been trying to get the img tag to
> work on my output.php page:
>
> Scout Photo:
> echo$row['photoName']; ?> "\nScout Photo : "." "\">"; ?> src=\"image.php?filename=".$row['photoName']."\">\n"; ?>
>\n"; ?>
>
>
>
> Here is one way I was trying to get to work for the image.php page where
> the headers are supposed to work and don't
>
> include("inc/dbconn_open.php");
> error_reporting(E_ALL);
>
>
> //if (isset($_FILES['ePhoto'])){$ePhoto = $_FILES['ePhoto'];} else {$ePhoto
> ="";}
>
> $filename = $_GET['$filename'];
>
> $image = stripslashes($_REQUEST[photoName]);
>
> $sql = "SELECT ePhoto, photoName, photoType from eagleProjects WHERE
> photoName = $filename";
> $result=mysql_query($sql);
> $data = mysql_fetch_array ($result);
>
>
>
> $type = $data['photoType'];
> $name = $data['photoName'];
>
> header("Content-type: $type");
> header("Content-Disposition: attachment; filename=$name");
>
>
> echo $data["photoName"];
> echo $data["ePhoto"];
>
> exit;
>
Terin,
I just gave the same code to another person on another forum and he had no
issues getting that to work. If it does not work, you will need to ensure
that the value you pass to the show_image file is the same value that you
are using to look up the image. So if its an id number make sure you are
using the same id field to look up the image.
--
Bastien
Cat, the other other white meat
Re: [PHP] Re: Displaying images
Well I have tried Numerous scripts and ways and still can't get the image to
display.
I have echoed the file and have been able to get the gibberish image code to
display but not a real image, here is my full code if anyone wants to have a
look, I am going crossed eyed here.
These are just a few of the ways I have been trying to get the img tag to work
on my output.php page:
Scout Photo:Scout Photo : ".""; ?> \n"; ?>
\n"; ?>
Here is one way I was trying to get to work for the image.php page where the
headers are supposed to work and don't
include("inc/dbconn_open.php");
error_reporting(E_ALL);
//if (isset($_FILES['ePhoto'])){$ePhoto = $_FILES['ePhoto'];} else {$ePhoto
="";}
$filename = $_GET['$filename'];
$image = stripslashes($_REQUEST[photoName]);
$sql = "SELECT ePhoto, photoName, photoType from eagleProjects WHERE photoName
= $filename";
$result=mysql_query($sql);
$data = mysql_fetch_array ($result);
$type = $data['photoType'];
$name = $data['photoName'];
header("Content-type: $type");
header("Content-Disposition: attachment; filename=$name");
echo $data["photoName"];
echo $data["ePhoto"];
exit;
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Re: [PHP] Re: Displaying images
Miller, Terion wrote:
[..]
> header("Content-type: img/jpeg");
[..]
> This page isn't working and if I try to browse this page it wants to
> open it with an editor, it won't view in the browser.
>
> What am I doing wrong? Is it the code or the data?
"Content-type: img/jpeg" is undefined, the browser would not know what to
do with it. Try "image/jpeg" which is the official MIME type.
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Re: [PHP] Re: Displaying images
I suggest you use firebug https://addons.mozilla.org/en-US/firefox/addon/1843 coming back to your case.. 1. use the inspect function of firebug to check if the is displaying correctly. 2 then use the net tab of firebug to see if the image is actually downloaded from the server. 3. open the link in new tab to see image is actually displayed. if you are struck at case 1: some problem in the echo statement of your main php case 2: firebug actually tells you what is the mistake you did case 3: i suggest one more firefox addon: live http headers see if the content-type and content-length header are being sent This page isn't working and if I try to browse this page it wants to open it with an editor, it won't view in the browser. i assume that the address in the URL bar of your browser starts with file:// rather than http:// ? simply put are you double clicking the file or clicking open with firefox ?
Re: [PHP] Re: Displaying images
Shawn McKenzie wrote:
> Miller, Terion wrote:
>> Thanks for the suggestions everyone, I have this now, but still no image
>> showing up
>>
>> It is stored as a blob in the database.
>>
>> on the output page I am calling it like this:
>>
>>
>
> I was assuming that ePhoto was the actual binary data? If so then you
> need something else here like id or something. I just used filename as
> an example. If ePhoto is actually the filename then you are OK here,
> but not below.
>
>> Then on the image.php page I have this:
>>
>> > include("inc/dbconn_open.php");
>> error_reporting(E_ALL);
>>
>> //image.php
>> header("Content-type: img/jpeg");
>> $filename = $_GET['filename'];
>>
>> $sql = "SELECT ePhoto from eagleProjects WHERE $filename= ePhoto";
>
> Ummm... This can't work. What is ePhoto??? The data or the filename?
>
>> $result=mysql_query($sql);
>> $row = mysql_fetch_assoc ($result);
>>
>>
>>
>> ?>
>>
>>
>>
>> This page isn't working and if I try to browse this page it wants to open it
>> with an editor, it won't view in the browser.
>>
>> What am I doing wrong? Is it the code or the data?
>>
>> Thanks
>> Terion
>
>
I don't know what your fields are, but if ePhoto is the BLOB, and you
have an id field, then try this (assuming that you have selected id in
your query and it is an integer):
//image.php
$id = (int)$_GET['id'];
$sql = "SELECT ePhoto from eagleProjects WHERE id=$id";
--
Thanks!
-Shawn
http://www.spidean.com
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Re: [PHP] Re: Displaying images
Miller, Terion wrote:
>
> Thanks for the suggestions everyone, I have this now, but still no image
> showing up
>
> It is stored as a blob in the database.
>
> on the output page I am calling it like this:
>
>
I was assuming that ePhoto was the actual binary data? If so then you
need something else here like id or something. I just used filename as
an example. If ePhoto is actually the filename then you are OK here,
but not below.
>
> Then on the image.php page I have this:
>
> include("inc/dbconn_open.php");
> error_reporting(E_ALL);
>
> //image.php
> header("Content-type: img/jpeg");
> $filename = $_GET['filename'];
>
> $sql = "SELECT ePhoto from eagleProjects WHERE $filename= ePhoto";
Ummm... This can't work. What is ePhoto??? The data or the filename?
> $result=mysql_query($sql);
> $row = mysql_fetch_assoc ($result);
>
>
>
> ?>
>
>
>
> This page isn't working and if I try to browse this page it wants to open it
> with an editor, it won't view in the browser.
>
> What am I doing wrong? Is it the code or the data?
>
> Thanks
> Terion
--
Thanks!
-Shawn
http://www.spidean.com
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Re: [PHP] Re: Displaying images
Thanks for the suggestions everyone, I have this now, but still no image showing up It is stored as a blob in the database. on the output page I am calling it like this: Then on the image.php page I have this: This page isn't working and if I try to browse this page it wants to open it with an editor, it won't view in the browser. What am I doing wrong? Is it the code or the data? Thanks Terion -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Re: Displaying images
On Wed, 2009-05-27 at 15:14 -0400, Bastien Koert wrote:
> On Wed, May 27, 2009 at 3:05 PM, Ashley Sheridan
> wrote:
>
> > On Wed, 2009-05-27 at 14:03 -0400, Bastien Koert wrote:
> > > On Wed, May 27, 2009 at 1:51 PM, Miller, Terion <
> > > tmil...@springfi.gannett.com> wrote:
> > >
> > > >
> > > >
> > > >
> > > > On 5/27/09 12:49 PM, "Bastien Koert" wrote:
> > > >
> > > > On Wed, May 27, 2009 at 1:46 PM, Shawn McKenzie > > > >wrote:
> > > >
> > > > > Miller, Terion wrote:
> > > > > > I am trying to get an image to display but I get nothing if done
> > like
> > > > > this:
> > > > > >
> > > > > >
> > > > > > Scout Photo:
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > > If I just echo the field I do get the file name
> > > > > >
> > > > >
> > > > > Does the filename include the path? Does the image with said
> > filename
> > > > > actually exist in that path?
> > > > >
> > > > > --
> > > > > Thanks!
> > > > > -Shawn
> > > > > http://www.spidean.com
> > > > >
> > > > > --
> > > > > PHP General Mailing List (http://www.php.net/)
> > > > > To unsubscribe, visit: http://www.php.net/unsub.php
> > > > >
> > > > >
> > > > Is it an image from the db or a path to an image on the filesystem?
> > > >
> > > > --
> > > >
> > > > Bastien
> > > >
> > > > Cat, the other other white meat
> > > >
> > > >
> > > > Hi Bastien it is an image in the db.
> > > > Thanks
> > > > Terion
> > > >
> > >
> > > You can't display the image that way due to differeing headers in
> > required
> > > for html and images. Change your image code to call a page that
> > specializes
> > > in handling the images and the headers needed by changing to something
> > like
> > > this:
> > >
> > > echo "\n";
> > >
> > > where the show_image page calls the image out from the db and display it
> > as
> > > below
> > >
> > > > > require("conn.php");
> > > //check to see if the id is passed
> > > if(isset($_GET['id'])) {
> > > $id=$_GET['id'];
> > >
> > > //query the database to get the image and the filetype
> > > $query = "select bin_data, filetype from binary_data where id=$id";
> > >
> > > $result = mysql_query($query);
> > > $row = mysql_fetch_array($result);
> > > {
> > >$data = $row['bin_data'];
> > >$type = $row['filetype'];
> > > }
> > > if ($type=="pjpeg") $type = "jpeg"; //handle the ms jpeg
> > alternate
> > > format
> > > Header( "Content-type: $type");
> > > echo $data;
> > > }
> > > ?>
> > That's only true if the image is stored in the database as a BLOB or
> > somesuch. If the database is just storing the filename of the image,
> > then it may just need the correct directory path prefix. Go into your
> > browser, right-click and check to see what path and filename the browser
> > expects to find the image file. My guess is it's not the same as where
> > the image actually is, and you need to add the path before opening the
> > > http://domain) or a relative one, and relative paths can include the ../
> > directory if you need to move up the directory tree.
> >
> >
> > Ash
> > www.ashleysheridan.co.uk
> >
> >
> Terion stated it was in the db
>
The thread didn't follow on properly, my email client wasn't able to
join t by thread so I didn't see his message until after I'd replied,
sorry!
Ash
www.ashleysheridan.co.uk
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Re: [PHP] Re: Displaying images
On Wed, May 27, 2009 at 3:05 PM, Ashley Sheridan
wrote:
> On Wed, 2009-05-27 at 14:03 -0400, Bastien Koert wrote:
> > On Wed, May 27, 2009 at 1:51 PM, Miller, Terion <
> > tmil...@springfi.gannett.com> wrote:
> >
> > >
> > >
> > >
> > > On 5/27/09 12:49 PM, "Bastien Koert" wrote:
> > >
> > > On Wed, May 27, 2009 at 1:46 PM, Shawn McKenzie > > >wrote:
> > >
> > > > Miller, Terion wrote:
> > > > > I am trying to get an image to display but I get nothing if done
> like
> > > > this:
> > > > >
> > > > >
> > > > > Scout Photo:
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > > If I just echo the field I do get the file name
> > > > >
> > > >
> > > > Does the filename include the path? Does the image with said
> filename
> > > > actually exist in that path?
> > > >
> > > > --
> > > > Thanks!
> > > > -Shawn
> > > > http://www.spidean.com
> > > >
> > > > --
> > > > PHP General Mailing List (http://www.php.net/)
> > > > To unsubscribe, visit: http://www.php.net/unsub.php
> > > >
> > > >
> > > Is it an image from the db or a path to an image on the filesystem?
> > >
> > > --
> > >
> > > Bastien
> > >
> > > Cat, the other other white meat
> > >
> > >
> > > Hi Bastien it is an image in the db.
> > > Thanks
> > > Terion
> > >
> >
> > You can't display the image that way due to differeing headers in
> required
> > for html and images. Change your image code to call a page that
> specializes
> > in handling the images and the headers needed by changing to something
> like
> > this:
> >
> > echo "\n";
> >
> > where the show_image page calls the image out from the db and display it
> as
> > below
> >
> > > require("conn.php");
> > //check to see if the id is passed
> > if(isset($_GET['id'])) {
> > $id=$_GET['id'];
> >
> > //query the database to get the image and the filetype
> > $query = "select bin_data, filetype from binary_data where id=$id";
> >
> > $result = mysql_query($query);
> > $row = mysql_fetch_array($result);
> > {
> >$data = $row['bin_data'];
> >$type = $row['filetype'];
> > }
> > if ($type=="pjpeg") $type = "jpeg"; //handle the ms jpeg
> alternate
> > format
> > Header( "Content-type: $type");
> > echo $data;
> > }
> > ?>
> That's only true if the image is stored in the database as a BLOB or
> somesuch. If the database is just storing the filename of the image,
> then it may just need the correct directory path prefix. Go into your
> browser, right-click and check to see what path and filename the browser
> expects to find the image file. My guess is it's not the same as where
> the image actually is, and you need to add the path before opening the
> http://domain) or a relative one, and relative paths can include the ../
> directory if you need to move up the directory tree.
>
>
> Ash
> www.ashleysheridan.co.uk
>
>
Terion stated it was in the db
--
Bastien
Cat, the other other white meat
Re: [PHP] Re: Displaying images
On Wed, 2009-05-27 at 14:03 -0400, Bastien Koert wrote:
> On Wed, May 27, 2009 at 1:51 PM, Miller, Terion <
> tmil...@springfi.gannett.com> wrote:
>
> >
> >
> >
> > On 5/27/09 12:49 PM, "Bastien Koert" wrote:
> >
> > On Wed, May 27, 2009 at 1:46 PM, Shawn McKenzie > >wrote:
> >
> > > Miller, Terion wrote:
> > > > I am trying to get an image to display but I get nothing if done like
> > > this:
> > > >
> > > >
> > > > Scout Photo:
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > > If I just echo the field I do get the file name
> > > >
> > >
> > > Does the filename include the path? Does the image with said filename
> > > actually exist in that path?
> > >
> > > --
> > > Thanks!
> > > -Shawn
> > > http://www.spidean.com
> > >
> > > --
> > > PHP General Mailing List (http://www.php.net/)
> > > To unsubscribe, visit: http://www.php.net/unsub.php
> > >
> > >
> > Is it an image from the db or a path to an image on the filesystem?
> >
> > --
> >
> > Bastien
> >
> > Cat, the other other white meat
> >
> >
> > Hi Bastien it is an image in the db.
> > Thanks
> > Terion
> >
>
> You can't display the image that way due to differeing headers in required
> for html and images. Change your image code to call a page that specializes
> in handling the images and the headers needed by changing to something like
> this:
>
> echo "\n";
>
> where the show_image page calls the image out from the db and display it as
> below
>
> require("conn.php");
> //check to see if the id is passed
> if(isset($_GET['id'])) {
> $id=$_GET['id'];
>
> //query the database to get the image and the filetype
> $query = "select bin_data, filetype from binary_data where id=$id";
>
> $result = mysql_query($query);
> $row = mysql_fetch_array($result);
> {
>$data = $row['bin_data'];
>$type = $row['filetype'];
> }
> if ($type=="pjpeg") $type = "jpeg"; //handle the ms jpeg alternate
> format
> Header( "Content-type: $type");
> echo $data;
> }
> ?>
That's only true if the image is stored in the database as a BLOB or
somesuch. If the database is just storing the filename of the image,
then it may just need the correct directory path prefix. Go into your
browser, right-click and check to see what path and filename the browser
expects to find the image file. My guess is it's not the same as where
the image actually is, and you need to add the path before opening the
http://domain) or a relative one, and relative paths can include the ../
directory if you need to move up the directory tree.
Ash
www.ashleysheridan.co.uk
--
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Re: [PHP] Re: Displaying images
Miller, Terion wrote:
>
>
>
> Does the filename include the path? Does the image with said filename
> actually exist in that path?
>
> --
> Thanks!
> -Shawn
> http://www.spidean.com
>
> --
> PHP General Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>
>
>
> Hmm guess I am confused because the image is being stored in the db.
> So I thought I could call it like any other field data.
>
> Thanks
> Terion
>
>
Please reply all.
The img tag is used to tell the browser where the image is located, it
then loads the image. If you have the binary image data in the DB then
you probably need to have a PHP file with a get var as the img src and
then have that PHP file actually output the image data:
//image.php
header("Content-type: img/jpeg");
$filename = $_GET['filename'];
//do your SQL select and fetch using filename or whatnot to get the
image data, I assume from 'ePhoto'
echo $row['ePhoto'];
Thanks!
-Shawn
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Re: [PHP] Re: Displaying images
On Wed, May 27, 2009 at 1:51 PM, Miller, Terion < tmil...@springfi.gannett.com> wrote: > > > > On 5/27/09 12:49 PM, "Bastien Koert" wrote: > > On Wed, May 27, 2009 at 1:46 PM, Shawn McKenzie >wrote: > > > Miller, Terion wrote: > > > I am trying to get an image to display but I get nothing if done like > > this: > > > > > > > > > Scout Photo: > > > > > > > > > > > > > > > > > > > > > If I just echo the field I do get the file name > > > > > > > Does the filename include the path? Does the image with said filename > > actually exist in that path? > > > > -- > > Thanks! > > -Shawn > > http://www.spidean.com > > > > -- > > PHP General Mailing List (http://www.php.net/) > > To unsubscribe, visit: http://www.php.net/unsub.php > > > > > Is it an image from the db or a path to an image on the filesystem? > > -- > > Bastien > > Cat, the other other white meat > > > Hi Bastien it is an image in the db. > Thanks > Terion > You can't display the image that way due to differeing headers in required for html and images. Change your image code to call a page that specializes in handling the images and the headers needed by changing to something like this: echo "\n"; where the show_image page calls the image out from the db and display it as below -- Bastien Cat, the other other white meat
Re: [PHP] Re: Displaying images
On 5/27/09 12:49 PM, "Bastien Koert" wrote: On Wed, May 27, 2009 at 1:46 PM, Shawn McKenzie wrote: > Miller, Terion wrote: > > I am trying to get an image to display but I get nothing if done like > this: > > > > > > Scout Photo: > > > > > > > > > > > > > > If I just echo the field I do get the file name > > > > Does the filename include the path? Does the image with said filename > actually exist in that path? > > -- > Thanks! > -Shawn > http://www.spidean.com > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > > Is it an image from the db or a path to an image on the filesystem? -- Bastien Cat, the other other white meat Hi Bastien it is an image in the db. Thanks Terion -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Re: Displaying images
On Wed, May 27, 2009 at 1:46 PM, Shawn McKenzie wrote: > Miller, Terion wrote: > > I am trying to get an image to display but I get nothing if done like > this: > > > > > > Scout Photo: > > > > > > > > > > > > > > If I just echo the field I do get the file name > > > > Does the filename include the path? Does the image with said filename > actually exist in that path? > > -- > Thanks! > -Shawn > http://www.spidean.com > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > > Is it an image from the db or a path to an image on the filesystem? -- Bastien Cat, the other other white meat
[PHP] Re: Displaying images
Miller, Terion wrote: > I am trying to get an image to display but I get nothing if done like this: > > > Scout Photo: > > > > > > > If I just echo the field I do get the file name > Does the filename include the path? Does the image with said filename actually exist in that path? -- Thanks! -Shawn http://www.spidean.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
