Miller, Terion wrote:
>
>
> 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 $data["ePhoto"];
>
> exit;
It's hard to tell, not knowing what the db fields contain, but the two
glaring issues are:
1. unless photoType is image/something then it won't work. It can't
just be jpeg or gif. So maybe do header("Content-type: image/$type");
assuming $type is correct.
2. Since $data["photoName"]; is not of type image/whatever, the echo may
break the display.
--
Thanks!
-Shawn
http://www.spidean.com
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