subject:"\[PHP\] Unable to display images on browser"

RE: [PHP] Unable to display images on browser

2002-03-01 Thread Narvaez, Teresa


Hello, 
Thanks for your responses; however, I do not think I have the gd
libraries installed because ImageCreateFromString() was not found.  I will
install it. 

I have a question: I can display the PNG or GIF image using the
browser.  So Why do I need the GD library?  Since I store the mime type in
the database I think that a call to header() to tell the browser what type
of mime-type is coming from the database should sufice.  Also, I want to
store any type of binary data into my database (PNG, GIF, word, JPG, etc) Am
I missing something?  

Thanks, 
-Teresa


-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
Sent: Thursday, February 28, 2002 4:12 AM
To: [EMAIL PROTECTED]
Subject: RE: [PHP] Unable to display images on browser


You need to insert the following lines after this line:
Header(Content-type: image/gif);

$im = ImageCreateFromString ($fileContent); 
ImageGif ($im); 

and then remove this line: echo $fileContent;
That should do it.

/Joakim

-Original Message-
From: Narvaez, Teresa [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, February 27, 2002 8:32 PM
To: '[EMAIL PROTECTED]'; [EMAIL PROTECTED]
Subject: RE: [PHP] Unable to display images on browser


Hello, Thanks for your help.  This is what I have for file1.php and
ddownloadfile.php.  What I want is to click on Donwnload now link and be
able to get the file out of the database and display it on the browser.
Thank you in adavance, -Teresa
file1.php 
--- 
?php 
   while ($row = mysql_fetch_array($result)) 
   { 
? 
  SOME_HTLM_CODE_TO_DISPLAY_DB_FIELDS_GOES_HERE; 
//  img src=\ddownloadfile.php?fileId=?php echo $row[PicNum]; ?\
 
  a href=ddownloadfile.php?fileId=?php echo $row[PicNum]; ?  
 Download Now 
  /a/font 
  /td 
  /tr 
?php


ddownloadfile.php 
-- 
?  
$dbQuery = Select PicNum, size, type, description, Image; 
$dbQuery .=  FROM Images WHERE PicNum = $fileId; 
$result = mysql_query($dbQuery) 
or die (Could not get file list:  . mysql_error() ); 
echo Sent Query successfullybr; 
if ( mysql_num_rows($result) == 1) 
{ 
$fileType = @mysql_result($result,0, type); 
$fileContent = @mysql_result($result, 0, Image); 
$filedesc = @mysql_result($result,0, description); 
$filenum = @mysql_result($result,0, PicNum); 
Header(Content-type: image/gif); 
echo $fileContent; 
} 
else 
{ 
echo Record does not exist; 
} // else 
? 

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Re: [PHP] Unable to display images on browser

2002-03-01 Thread Andrew Brampton


if u have the data stored in the DB, just chuck the data out, with the
correct mime-type header..

Andrew
- Original Message -
From: Narvaez, Teresa [EMAIL PROTECTED]
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Friday, March 01, 2002 8:41 PM
Subject: RE: [PHP] Unable to display images on browser


 Hello,
 Thanks for your responses; however, I do not think I have the gd
 libraries installed because ImageCreateFromString() was not found.  I will
 install it.

 I have a question: I can display the PNG or GIF image using the
 browser.  So Why do I need the GD library?  Since I store the mime type in
 the database I think that a call to header() to tell the browser what type
 of mime-type is coming from the database should sufice.  Also, I want to
 store any type of binary data into my database (PNG, GIF, word, JPG, etc)
Am
 I missing something?

 Thanks,
 -Teresa


 -Original Message-
 From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
 Sent: Thursday, February 28, 2002 4:12 AM
 To: [EMAIL PROTECTED]
 Subject: RE: [PHP] Unable to display images on browser


 You need to insert the following lines after this line:
 Header(Content-type: image/gif);

 $im = ImageCreateFromString ($fileContent);
 ImageGif ($im);

 and then remove this line: echo $fileContent;
 That should do it.

 /Joakim

 -Original Message-
 From: Narvaez, Teresa [mailto:[EMAIL PROTECTED]]
 Sent: Wednesday, February 27, 2002 8:32 PM
 To: '[EMAIL PROTECTED]'; [EMAIL PROTECTED]
 Subject: RE: [PHP] Unable to display images on browser


 Hello, Thanks for your help.  This is what I have for file1.php and
 ddownloadfile.php.  What I want is to click on Donwnload now link and be
 able to get the file out of the database and display it on the browser.
 Thank you in adavance, -Teresa
 file1.php
 ---
 ?php
while ($row = mysql_fetch_array($result))
{
 ?
   SOME_HTLM_CODE_TO_DISPLAY_DB_FIELDS_GOES_HERE;
 //  img src=\ddownloadfile.php?fileId=?php echo $row[PicNum];
?\
 
   a href=ddownloadfile.php?fileId=?php echo $row[PicNum]; ? 
  Download Now
   /a/font
   /td
   /tr
 ?php


 ddownloadfile.php
 --
 ?
 $dbQuery = Select PicNum, size, type, description, Image;
 $dbQuery .=  FROM Images WHERE PicNum = $fileId;
 $result = mysql_query($dbQuery)
 or die (Could not get file list:  . mysql_error() );
 echo Sent Query successfullybr;
 if ( mysql_num_rows($result) == 1)
 {
 $fileType = @mysql_result($result,0, type);
 $fileContent = @mysql_result($result, 0, Image);
 $filedesc = @mysql_result($result,0, description);
 $filenum = @mysql_result($result,0, PicNum);
 Header(Content-type: image/gif);
 echo $fileContent;
 }
 else
 {
 echo Record does not exist;
 } // else
 ?

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RE: [PHP] Unable to display images on browser

2002-03-01 Thread Narvaez, Teresa


Andrew,
That is exactly what I am trying to do but the only thing I get is a
box with an X on IE 4.0 and Netscape Communicator 4.73.  Please see my a
fragement of my php code below. Thanks in advance! -Teresa

--
file1.php
--
?php
   while ($row = mysql_fetch_array($result))
   {
?
  SOME_HTLM_CODE_TO_DISPLAY_DB_FIELDS_GOES_HERE;
  a href=ddownloadfile.php?fileId=?php echo $row[PicNum]; ? 
 Download Now
?php 
   } 
?
--
ddownloadfile.php
--
?
$dbQuery = Select PicNum, size, type, description, Image;
$dbQuery .=  FROM Images WHERE PicNum = $fileId;
$result = mysql_query($dbQuery)
or die (Could not get file list:  . mysql_error() );
echo Sent Query successfullybr;
if ( mysql_num_rows($result) == 1)
{
  $fileType = @mysql_result($result,0, type);
  $fileContent = @mysql_result($result, 0, Image);
  $filedesc = @mysql_result($result,0, description);
  $filenum = @mysql_result($result,0, PicNum);
// Header(Content-type: image/gif);
  Header(Content-type: $fileType);
  echo $fileContent;
}
else
{
  echo Record does not exist;
} // else

?
-Original Message-
From: Andrew Brampton [mailto:[EMAIL PROTECTED]]
Sent: Friday, March 01, 2002 5:08 PM
To: Narvaez, Teresa; [EMAIL PROTECTED];
[EMAIL PROTECTED]
Subject: Re: [PHP] Unable to display images on browser


if u have the data stored in the DB, just chuck the data out, with the
correct mime-type header..

Andrew


--


- Original Message -
From: Narvaez, Teresa [EMAIL PROTECTED]
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Friday, March 01, 2002 8:41 PM
Subject: RE: [PHP] Unable to display images on browser


 Hello,
 Thanks for your responses; however, I do not think I have the gd
 libraries installed because ImageCreateFromString() was not found.  I will
 install it.

 I have a question: I can display the PNG or GIF image using the
 browser.  So Why do I need the GD library?  Since I store the mime type in
 the database I think that a call to header() to tell the browser what type
 of mime-type is coming from the database should sufice.  Also, I want to
 store any type of binary data into my database (PNG, GIF, word, JPG, etc)
Am
 I missing something?

 Thanks,
 -Teresa


 -Original Message-
 From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
 Sent: Thursday, February 28, 2002 4:12 AM
 To: [EMAIL PROTECTED]
 Subject: RE: [PHP] Unable to display images on browser


 You need to insert the following lines after this line:
 Header(Content-type: image/gif);

 $im = ImageCreateFromString ($fileContent);
 ImageGif ($im);

 and then remove this line: echo $fileContent;
 That should do it.

 /Joakim

 -Original Message-
 From: Narvaez, Teresa [mailto:[EMAIL PROTECTED]]
 Sent: Wednesday, February 27, 2002 8:32 PM
 To: '[EMAIL PROTECTED]'; [EMAIL PROTECTED]
 Subject: RE: [PHP] Unable to display images on browser


 Hello, Thanks for your help.  This is what I have for file1.php and
 ddownloadfile.php.  What I want is to click on Donwnload now link and be
 able to get the file out of the database and display it on the browser.
 Thank you in adavance, -Teresa
 file1.php
 ---
 ?php
while ($row = mysql_fetch_array($result))
{
 ?
   SOME_HTLM_CODE_TO_DISPLAY_DB_FIELDS_GOES_HERE;
 //  img src=\ddownloadfile.php?fileId=?php echo $row[PicNum];
?\
 
   a href=ddownloadfile.php?fileId=?php echo $row[PicNum]; ? 
  Download Now
   /a/font
   /td
   /tr
 ?php


 ddownloadfile.php
 --
 ?
 $dbQuery = Select PicNum, size, type, description, Image;
 $dbQuery .=  FROM Images WHERE PicNum = $fileId;
 $result = mysql_query($dbQuery)
 or die (Could not get file list:  . mysql_error() );
 echo Sent Query successfullybr;
 if ( mysql_num_rows($result) == 1)
 {
 $fileType = @mysql_result($result,0, type);
 $fileContent = @mysql_result($result, 0, Image);
 $filedesc = @mysql_result($result,0, description);
 $filenum = @mysql_result($result,0, PicNum);
 Header(Content-type: image/gif);
 echo $fileContent;
 }
 else
 {
 echo Record does not exist;
 } // else
 ?

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Re: [PHP] Unable to display images on browser

2002-03-01 Thread Andrew Brampton


RE: [PHP] Unable to display images on browserI've never done this in PHP, but one way 
of testing if your output is correct is telnet to your webserver and request the php 
script yourself. Then request a valid JPG (or whatever), and compare the outputs..

If you have netcat (great tool), you can do this:
nc localhost 80 -v  output.txt
GET /ddownloadfile.php?fileId=1 HTTP/1.0

and in output.txt would be some headers and some raw data, and u can compare this to a 
normal jpg request.

If you don't have netcat, then just use standard telnet that comes with windows and 
type 
GET /ddownloadfile.php?fileId=1 HTTP/1.0
for the request as normal

Hope this helps
Andrew
  - Original Message - 
  From: Narvaez, Teresa 
  To: 'Andrew Brampton' ; [EMAIL PROTECTED] ; [EMAIL PROTECTED] 
  Sent: Friday, March 01, 2002 10:21 PM
  Subject: RE: [PHP] Unable to display images on browser


  Andrew, 
  That is exactly what I am trying to do but the only thing I get is a box 
with an X on IE 4.0 and Netscape Communicator 4.73.  Please see my a fragement of my 
php code below. Thanks in advance! -Teresa

  -- 
  file1.php 
  -- 
  ?php 
 while ($row = mysql_fetch_array($result)) 
 { 
  ? 
SOME_HTLM_CODE_TO_DISPLAY_DB_FIELDS_GOES_HERE; 
a href=ddownloadfile.php?fileId=?php echo $row[PicNum]; ?  
   Download Now 
  ?php 
 } 
  ? 
  -- 
  ddownloadfile.php 
  -- 
  ? 
  $dbQuery = Select PicNum, size, type, description, Image; 
  $dbQuery .=  FROM Images WHERE PicNum = $fileId; 
  $result = mysql_query($dbQuery) 
  or die (Could not get file list:  . mysql_error() ); 
  echo Sent Query successfullybr; 
  if ( mysql_num_rows($result) == 1) 
  { 
$fileType = @mysql_result($result,0, type); 
$fileContent = @mysql_result($result, 0, Image); 
$filedesc = @mysql_result($result,0, description); 
$filenum = @mysql_result($result,0, PicNum); 
  // Header(Content-type: image/gif); 
Header(Content-type: $fileType); 
echo $fileContent; 
  } 
  else 
  { 
echo Record does not exist; 
  } // else 

  ? 
  -Original Message- 
  From: Andrew Brampton [mailto:[EMAIL PROTECTED]] 
  Sent: Friday, March 01, 2002 5:08 PM 
  To: Narvaez, Teresa; [EMAIL PROTECTED]; 
  [EMAIL PROTECTED] 
  Subject: Re: [PHP] Unable to display images on browser 



  if u have the data stored in the DB, just chuck the data out, with the 
  correct mime-type header.. 

  Andrew 



  -- 



  - Original Message - 
  From: Narvaez, Teresa [EMAIL PROTECTED] 
  To: [EMAIL PROTECTED]; [EMAIL PROTECTED] 
  Sent: Friday, March 01, 2002 8:41 PM 
  Subject: RE: [PHP] Unable to display images on browser 



   Hello, 
   Thanks for your responses; however, I do not think I have the gd 
   libraries installed because ImageCreateFromString() was not found.  I will 
   install it. 
   
   I have a question: I can display the PNG or GIF image using the 
   browser.  So Why do I need the GD library?  Since I store the mime type in 
   the database I think that a call to header() to tell the browser what type 
   of mime-type is coming from the database should sufice.  Also, I want to 
   store any type of binary data into my database (PNG, GIF, word, JPG, etc) 
  Am 
   I missing something? 
   
   Thanks, 
   -Teresa 
   
   
   -Original Message- 
   From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]] 
   Sent: Thursday, February 28, 2002 4:12 AM 
   To: [EMAIL PROTECTED] 
   Subject: RE: [PHP] Unable to display images on browser 
   
   
   You need to insert the following lines after this line: 
   Header(Content-type: image/gif); 
   
   $im = ImageCreateFromString ($fileContent); 
   ImageGif ($im); 
   
   and then remove this line: echo $fileContent; 
   That should do it. 
   
   /Joakim 
   
   -Original Message- 
   From: Narvaez, Teresa [mailto:[EMAIL PROTECTED]] 
   Sent: Wednesday, February 27, 2002 8:32 PM 
   To: '[EMAIL PROTECTED]'; [EMAIL PROTECTED] 
   Subject: RE: [PHP] Unable to display images on browser 
   
   
   Hello, Thanks for your help.  This is what I have for file1.php and 
   ddownloadfile.php.  What I want is to click on Donwnload now link and be 
   able to get the file out of the database and display it on the browser. 
   Thank you in adavance, -Teresa 
   file1.php 
   --- 
   ?php 
  while ($row = mysql_fetch_array($result)) 
  { 
   ? 
 SOME_HTLM_CODE_TO_DISPLAY_DB_FIELDS_GOES_HERE; 
   //  img src=\ddownloadfile.php?fileId=?php echo $row[PicNum]; 
  ?\ 

 a href=ddownloadfile.php?fileId=?php echo $row[PicNum]; ?  
Download Now 
 /a/font 
 /td 
 /tr 
   ?php 
   
   
   ddownloadfile.php 
   -- 
   ? 
   $dbQuery = Select PicNum, size, type, description, Image; 
   $dbQuery

RE: [PHP] Unable to display images on browser

2002-02-28 Thread joakim . andersson


You need to insert the following lines after this line:
Header(Content-type: image/gif);

$im = ImageCreateFromString ($fileContent); 
ImageGif ($im); 

and then remove this line: echo $fileContent;
That should do it.

/Joakim

-Original Message-
From: Narvaez, Teresa [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, February 27, 2002 8:32 PM
To: '[EMAIL PROTECTED]'; [EMAIL PROTECTED]
Subject: RE: [PHP] Unable to display images on browser


Hello, Thanks for your help.  This is what I have for file1.php and
ddownloadfile.php.  What I want is to click on Donwnload now link and be
able to get the file out of the database and display it on the browser.
Thank you in adavance, -Teresa
file1.php 
--- 
?php 
   while ($row = mysql_fetch_array($result)) 
   { 
? 
  SOME_HTLM_CODE_TO_DISPLAY_DB_FIELDS_GOES_HERE; 
//  img src=\ddownloadfile.php?fileId=?php echo $row[PicNum]; ?\
 
  a href=ddownloadfile.php?fileId=?php echo $row[PicNum]; ?  
 Download Now 
  /a/font 
  /td 
  /tr 
?php


ddownloadfile.php 
-- 
?  
$dbQuery = Select PicNum, size, type, description, Image; 
$dbQuery .=  FROM Images WHERE PicNum = $fileId; 
$result = mysql_query($dbQuery) 
or die (Could not get file list:  . mysql_error() ); 
echo Sent Query successfullybr; 
if ( mysql_num_rows($result) == 1) 
{ 
$fileType = @mysql_result($result,0, type); 
$fileContent = @mysql_result($result, 0, Image); 
$filedesc = @mysql_result($result,0, description); 
$filenum = @mysql_result($result,0, PicNum); 
Header(Content-type: image/gif); 
echo $fileContent; 
} 
else 
{ 
echo Record does not exist; 
} // else 
? 

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Re: [PHP] Unable to display images on browser

2002-02-27 Thread Stewart G.


try using gd if you have = 1.8.3 then you will have to use ImagePng() and 
Content-type: image/png instead of ImageGif() to export the image.

header (Content-type: image/gif);
$im = ImageCreateFromString ($fileContent);
ImageGif ($im);

=S.

On Tue, 26 Feb 2002, Narvaez, Teresa wrote:

 Hello, 
   I do not understand why I can't display images retrieved from MySQL
 on my browser(IE 4.0).  When I retrieve the image from MYSQL I set the
 Header function to change the type of content(image/gif) I am sending to the
 browser. However, the browser displays an box with an X in it.  I would
 greatly appretiate any ideas/help.  Here is the piece of code:  Thanks in
 advance! -Teresa
 
 ?
  Code to connect and selected DB not included
 -
 $dbQuery = Select PicNum, size, type, description, Image;
 $dbQuery .=  FROM Images WHERE PicNum = $fileId;
 
 $result = mysql_query($dbQuery)
 or die (Could not get file list:  . mysql_error() );
 echo Sent Query successfullybr;
 
 if ( mysql_num_rows($result) == 1)
 {
 $fileType = @mysql_result($result,0, type);
 $fileContent = @mysql_result($result, 0, Image);
 $filedesc = @mysql_result($result,0, description);
 $filenum = @mysql_result($result,0, PicNum);
 //  Header(Content-type: $fileType); 
 Header(Content-type: image/gif);
 echo $fileContent;
 }   
 else
 {
 echo Record does not exist;
 } // else
 ? 
 


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RE: [PHP] Unable to display images on browser

2002-02-27 Thread joakim . andersson


Well, you probably need to do something like this:

file1.php (where you want to display an image)

img src=getimage.php?fileId=42

end file1


getimage.php (the ? needs to be on the first line in the file)
?
  Code to connect and selected DB not included
 -
 $dbQuery = Select PicNum, size, type, description, Image;
 $dbQuery .=  FROM Images WHERE PicNum = $fileId;
 
 $result = mysql_query($dbQuery)
 or die (Could not get file list:  . mysql_error() );

 if ( mysql_num_rows($result) == 1)
 {
 $fileType = @mysql_result($result,0, type);
 $fileContent = @mysql_result($result, 0, Image);
 $filedesc = @mysql_result($result,0, description);
 $filenum = @mysql_result($result,0, PicNum);
 //  Header(Content-type: $fileType); 

header (Content-type: image/gif);
$im = ImageCreateFromString ($fileContent);
ImageGif ($im);

 }   
 else
 {
   //Insert code to display error.gif
 } // else
? 
---end getimage.php

Good Luck
/Joakim


 -Original Message-
 From: Narvaez, Teresa [mailto:[EMAIL PROTECTED]]
 Sent: Wednesday, February 27, 2002 2:08 AM
 To: 'Dean Householder'; [EMAIL PROTECTED]
 Subject: [PHP] Unable to display images on browser
 
 
 Hello, 
   I do not understand why I can't display images 
 retrieved from MySQL
 on my browser(IE 4.0).  When I retrieve the image from MYSQL I set the
 Header function to change the type of content(image/gif) I am 
 sending to the
 browser. However, the browser displays an box with an X in 
 it.  I would
 greatly appretiate any ideas/help.  Here is the piece of 
 code:  Thanks in
 advance! -Teresa
 
 ?
  Code to connect and selected DB not included
 -
 $dbQuery = Select PicNum, size, type, description, Image;
 $dbQuery .=  FROM Images WHERE PicNum = $fileId;
 
 $result = mysql_query($dbQuery)
 or die (Could not get file list:  . mysql_error() );
 echo Sent Query successfullybr;
 
 if ( mysql_num_rows($result) == 1)
 {
 $fileType = @mysql_result($result,0, type);
 $fileContent = @mysql_result($result, 0, Image);
 $filedesc = @mysql_result($result,0, description);
 $filenum = @mysql_result($result,0, PicNum);
 //  Header(Content-type: $fileType); 
 Header(Content-type: image/gif);
 echo $fileContent;
 }   
 else
 {
 echo Record does not exist;
 } // else
 ? 
 

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RE: [PHP] Unable to display images on browser

2002-02-27 Thread Narvaez, Teresa


Hello, Thanks for your help.  This is what I have for file1.php and
ddownloadfile.php.  What I want is to click on Donwnload now link and be
able to get the file out of the database and display it on the browser.
Thank you in adavance, -Teresa

file1.php
---
?php

   while ($row = mysql_fetch_array($result))
   {
? 
  SOME_HTLM_CODE_TO_DISPLAY_DB_FIELDS_GOES_HERE; 

//  img src=\ddownloadfile.php?fileId=?php echo $row[PicNum]; ?\

  a href=ddownloadfile.php?fileId=?php echo $row[PicNum]; ? 
 Download Now
  /a/font
  /td
  /tr

?php


ddownloadfile.php
--
?  
$dbQuery = Select PicNum, size, type, description, Image;
$dbQuery .=  FROM Images WHERE PicNum = $fileId;

$result = mysql_query($dbQuery)
or die (Could not get file list:  . mysql_error() );
echo Sent Query successfullybr;

if ( mysql_num_rows($result) == 1)
{
$fileType = @mysql_result($result,0, type);
$fileContent = @mysql_result($result, 0, Image);
$filedesc = @mysql_result($result,0, description);
$filenum = @mysql_result($result,0, PicNum);
Header(Content-type: image/gif);
echo $fileContent;
}
else
{
echo Record does not exist;
} // else

? 

-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, February 27, 2002 4:56 AM
To: [EMAIL PROTECTED]
Subject: RE: [PHP] Unable to display images on browser


Well, you probably need to do something like this:

file1.php (where you want to display an image)

img src=getimage.php?fileId=42

end file1


getimage.php (the ? needs to be on the first line in the file)
?
  Code to connect and selected DB not included
 -
 $dbQuery = Select PicNum, size, type, description, Image;
 $dbQuery .=  FROM Images WHERE PicNum = $fileId;
 
 $result = mysql_query($dbQuery)
 or die (Could not get file list:  . mysql_error() );

 if ( mysql_num_rows($result) == 1)
 {
 $fileType = @mysql_result($result,0, type);
 $fileContent = @mysql_result($result, 0, Image);
 $filedesc = @mysql_result($result,0, description);
 $filenum = @mysql_result($result,0, PicNum);
 //  Header(Content-type: $fileType); 

header (Content-type: image/gif);
$im = ImageCreateFromString ($fileContent);
ImageGif ($im);

 }   
 else
 {
   //Insert code to display error.gif
 } // else
? 
---end getimage.php

Good Luck
/Joakim


 -Original Message-
 From: Narvaez, Teresa [mailto:[EMAIL PROTECTED]]
 Sent: Wednesday, February 27, 2002 2:08 AM
 To: 'Dean Householder'; [EMAIL PROTECTED]
 Subject: [PHP] Unable to display images on browser
 
 
 Hello, 
   I do not understand why I can't display images 
 retrieved from MySQL
 on my browser(IE 4.0).  When I retrieve the image from MYSQL I set the
 Header function to change the type of content(image/gif) I am 
 sending to the
 browser. However, the browser displays an box with an X in 
 it.  I would
 greatly appretiate any ideas/help.  Here is the piece of 
 code:  Thanks in
 advance! -Teresa
 
 ?
  Code to connect and selected DB not included
 -
 $dbQuery = Select PicNum, size, type, description, Image;
 $dbQuery .=  FROM Images WHERE PicNum = $fileId;
 
 $result = mysql_query($dbQuery)
 or die (Could not get file list:  . mysql_error() );
 echo Sent Query successfullybr;
 
 if ( mysql_num_rows($result) == 1)
 {
 $fileType = @mysql_result($result,0, type);
 $fileContent = @mysql_result($result, 0, Image);
 $filedesc = @mysql_result($result,0, description);
 $filenum = @mysql_result($result,0, PicNum);
 //  Header(Content-type: $fileType); 
 Header(Content-type: image/gif);
 echo $fileContent;
 }   
 else
 {
 echo Record does not exist;
 } // else
 ? 
 

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[PHP] Unable to display images on browser

2002-02-26 Thread Narvaez, Teresa


Hello, 
I do not understand why I can't display images retrieved from MySQL
on my browser(IE 4.0).  When I retrieve the image from MYSQL I set the
Header function to change the type of content(image/gif) I am sending to the
browser. However, the browser displays an box with an X in it.  I would
greatly appretiate any ideas/help.  Here is the piece of code:  Thanks in
advance! -Teresa

?
 Code to connect and selected DB not included
-
$dbQuery = Select PicNum, size, type, description, Image;
$dbQuery .=  FROM Images WHERE PicNum = $fileId;

$result = mysql_query($dbQuery)
or die (Could not get file list:  . mysql_error() );
echo Sent Query successfullybr;

if ( mysql_num_rows($result) == 1)
{
$fileType = @mysql_result($result,0, type);
$fileContent = @mysql_result($result, 0, Image);
$filedesc = @mysql_result($result,0, description);
$filenum = @mysql_result($result,0, PicNum);
//  Header(Content-type: $fileType); 
Header(Content-type: image/gif);
echo $fileContent;
}   
else
{
echo Record does not exist;
} // else
?

RE: [PHP] Unable to display images on browser

Re: [PHP] Unable to display images on browser

RE: [PHP] Unable to display images on browser

Re: [PHP] Unable to display images on browser

RE: [PHP] Unable to display images on browser

Re: [PHP] Unable to display images on browser

RE: [PHP] Unable to display images on browser

RE: [PHP] Unable to display images on browser

[PHP] Unable to display images on browser

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