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 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: $fileType");
> Header("Content-type: image/gif");
> echo $fileContent;
> }
> else
> {
> echo "Record does not exist";
> } // else
> ?>
>
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