You'll need to create a separate php page called image.php or something and retrieve the $mydata2->photo and output it to the browser.
<img src='image.php?id=1'> image.php --- $sql = "select where id = '$id'"; echo $mydata2->photo; Nate -----Original Message----- From: John Taylor-Johnston [mailto:[EMAIL PROTECTED] Sent: Friday, November 12, 2004 9:29 PM To: [EMAIL PROTECTED] Cc: John Taylor-Johnston Subject: [PHP] Display an image I have a longblob with a jpeg loaded in it. But how do I display it? `photo` longblob NOT NULL, This doesn't work :) I need header information, etc. So what does? echo "<img src=".$mydata2->photo.">\n"; I don't have an example to work with. John ----------snip------------- $sql2 = "select * from ".$db.".".$table"; $news2 = mysql_query($sql2); echo "<img src=".$mydata2->photo.">\n"; echo "<h3>$mydata2->Rank $mydata2->Name</h3>\n"; echo "<small><i>$mydata2->biography</i></small>\n"; } -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
