Posted by Jerry Stuckle on 07/25/06 02:04

Michael martin wrote:
> I'm trying to create a picture database for a site that I'm working on and I
> would like to do the following:
> - (main goal) Have a php script load an entire directory of photos into my
> mysql database in one swoop.
> - Display thumbnails on a general viewing page.
> - allow for viewing individual pictures by clicking on the thumbnails.
>
> <?php
> error_reporting(E_ALL);
> $id = $_REQUEST["iid"];
> $link = mysql_connect("localhost", "root", "") or die("Could not connect:
> " . mysql_error());
>
> mysql_select_db("mysql") or die(mysql_error());
> $sql = "SELECT b1 FROM t1 where id in (1,2,3)";
>
> $result = mysql_query("$sql") or die("Invalid query: " . mysql_error());
> header("Content-type: image/jpeg");
>
> while ($row = mysql_fetch_array($result))
> {
> $fileContent = $row['b1'];
>
> $im = imagecreatefromstring($fileContent);
> $width = imagesx($im);
> $height = imagesy($im);
> $imgw = 50;
> $imgh = $height / $width * $imgw;
> $thumb = ImageCreate($imgw,$imgh);
>
> ImageCopyResized($thumb,$im,0,0,0,0,$imgw,$imgh,ImageSX($im),ImageSY($im));
> ImagejpeG($thumb);
>
> imagedestroy ($im);
> imagedestroy ($thumb);
> mysql_close ($link);
> }
> ?>
>
> It is only displaying one image.
>
> Thanks
> M.
>
>

You can't do it this way. The content-type header indicates a single
image will be generated.

You can split this into 2 files - the first one gets the rows and calls
the second once for each row to display the jpeg.

--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attglobal.net
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