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 successfully<br>";
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 successfully<br>";
> 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
> ?>
>
> --
> PHP General Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>
