This is what I used to do what you're trying: header("Content-Type: image/" . $imagetype); // $imagetype is jpeg or gif header("Content-Length: " . strlen($image)); echo $image;
Note that "Type" is capitalized in "Content-Type", and include the "Content-Length" as well. -Steve On Thursday, December 6, 2001, at 02:45 PM, phantom wrote: > I have successfully placed images (jpg,gif,png) into a MySQL database > BLOB field, now I want to be able to pull the data out and diplay it. > > based on tutorial at > http://www.zdnet.com/devhead/stories/articles/0,4413,2644827,00.html > > after Querying the DB I have the following lines: > > $ImgFile = mysql_result($Results,0,ImgFile); \\ image data; > $ImgType = mysql_result($Results,0,ImgType); \\ image type (image/jpeg, > image/gif); > header("Content-type: ${ImgType}"); > echo "$ImgFile"; > > Problem: This method, when it pulls up an image, it shows no image but a > bunch of greek code. Example > http://www.phantomcougar.com/strawberrypie_com/mem/show_img.php?PNum=8 > ((be sure to view page source) > > > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, e-mail: [EMAIL PROTECTED] > For additional commands, e-mail: [EMAIL PROTECTED] > To contact the list administrators, e-mail: [EMAIL PROTECTED] > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]