RE: [PHP-DB] echo

try echo "<img src='/album/img/{$row["photoFileName"]}' /> "; warpping the array element in braces allows for proper evaluation bastien

From: elk dolk <elkdolk@yahoo.com> To: php-db@lists.php.net Subject: [PHP-DB] echo Date: Thu, 29 Mar 2007 05:08:36 -0700 (PDT) thanks to Chris and Dimiter, I think I am close but still the problem is not solved, when I add echo "<img src='/album/img/".$row["photoFileName"]."' /> to the code as it was sugested by Dimiter there is parse Error : PHP Parse error: syntax error, unexpected $end in C:\Inetpub\wwwroot\album\show.php on line 44 line 44 is end of the code just after </html> what does it mean? I am storing just the name of photos in the database and the photos are in /img folder ,and there is no permissions issue. My testing server is IIS And the path would be something like this : Inetpub\wwwroot\album\img as I am running out of time! could someone complete this code just with one echo and img src so that I can retrive my photos ? MySQL columns : photoID=seq number

               photoFileName=name of my photo like 3sw.jpg
               title=title
               description=short description

----------------------------------------------------------------- <body> <?php $link = mysql_connect('localhost', 'root', 'pw'); if (!$link) {

             die('Not connected : ' . mysql_error());

}

              echo 'connected!';

$db_selected = mysql_select_db('album', $link); if (!$db_selected) {

                    die ('Can\'t use foo : ' . mysql_error());

} $query = "SELECT * FROM photo"; $result=mysql_query($query); while ($row = mysql_fetch_array($result)) { echo "<img src='/album/img/".$row["photoFileName"]."' /> } mysql_free_result($result); ?> </body> </html> --------------------------------- Looking for earth-friendly autos? Browse Top Cars by "Green Rating" at Yahoo! Autos' Green Center.

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