On Wed, May 27, 2009 at 1:51 PM, Miller, Terion < tmiller@xxxxxxxxxxxxxxxxxxxx> wrote: > > > > On 5/27/09 12:49 PM, "Bastien Koert" <phpster@xxxxxxxxx> wrote: > > On Wed, May 27, 2009 at 1:46 PM, Shawn McKenzie <nospam@xxxxxxxxxxxxx > >wrote: > > > Miller, Terion wrote: > > > I am trying to get an image to display but I get nothing if done like > > this: > > > > > > <tr> > > > <td>Scout Photo:</td> > > > > > > <td><img src="<?php echo $row['ePhoto'];?>"></td> > > > > > > </tr> > > > > > > > > > If I just echo the field I do get the file name.... > > > > > > > Does the filename include the path? Does the image with said filename > > actually exist in that path? > > > > -- > > Thanks! > > -Shawn > > http://www.spidean.com > > > > -- > > PHP General Mailing List (http://www.php.net/) > > To unsubscribe, visit: http://www.php.net/unsub.php > > > > > Is it an image from the db or a path to an image on the filesystem? > > -- > > Bastien > > Cat, the other other white meat > > > Hi Bastien it is an image in the db. > Thanks > Terion > You can't display the image that way due to differeing headers in required for html and images. Change your image code to call a page that specializes in handling the images and the headers needed by changing to something like this: echo "<img src=\"show_image.php?id=".$rows['id']."\">\n"; where the show_image page calls the image out from the db and display it as below <?php require("conn.php"); //check to see if the id is passed if(isset($_GET['id'])) { $id=$_GET['id']; //query the database to get the image and the filetype $query = "select bin_data, filetype from binary_data where id=$id"; $result = mysql_query($query); $row = mysql_fetch_array($result); { $data = $row['bin_data']; $type = $row['filetype']; } if ($type=="pjpeg") $type = "jpeg"; //handle the ms jpeg alternate format Header( "Content-type: $type"); echo $data; } ?> -- Bastien Cat, the other other white meat
