Re: Re: Displaying images

Well I have tried Numerous scripts and ways and still can't get the image to display. I have echoed the file and have been able to get the gibberish image code to display but not a real image, here is my full code if anyone wants to have a look, I am going crossed eyed here. These are just a few of the ways I have been trying to get the img tag to work on my output.php page: <tr> <td>Scout Photo:</td> <td> <a href="image.php?filename=<?php echo$row['photoName']; ?>"><?php echo$row['photoName']; ?></A><br> <?php echo "\n<br>Scout Photo : "."<img src=\image.php?filename=" .$row['photoName']. "\">"; ?> <?php echo "<img src=\"image.php?filename=".$row['photoName']."\">\n"; ?> <?php echo "<img src=\"image.php?id=".$row['ePID']."\">\n"; ?> <img src="image2.php?filename=<?php echo $row['photoName']; ?>"> Here is one way I was trying to get to work for the image.php page where the headers are supposed to work and don't include("inc/dbconn_open.php"); error_reporting(E_ALL); //if (isset($_FILES['ePhoto'])){$ePhoto = $_FILES['ePhoto'];} else {$ePhoto ="";} $filename = $_GET['$filename']; $image = stripslashes($_REQUEST[photoName]); $sql = "SELECT ePhoto, photoName, photoType from eagleProjects WHERE photoName = $filename"; $result=mysql_query($sql); $data = mysql_fetch_array ($result); $type = $data['photoType']; $name = $data['photoName']; header("Content-type: $type"); header("Content-Disposition: attachment; filename=$name"); echo $data["photoName"]; echo $data["ePhoto"]; exit;