try

echo "<img src='/album/img/{$row["photoFileName"]}' /> ";

warpping the array element in braces allows for proper evaluation

bastien

From: elk dolk <[EMAIL PROTECTED]>
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>



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