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
