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
> ?>
>
> --
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>
