try
echo "<img src='/album/img/{$row["photoFileName"]}' /> ";
warpping the array element in braces allows for proper evaluation
bastien
From: elk dolk <elkdolk@xxxxxxxxx>
To: php-db@xxxxxxxxxxxxx
Subject: echo
Date: Thu, 29 Mar 2007 05:08:36 -0700 (PDT)
thanks to Chris and Dimiter,
I think I am close but still the problem is not solved, when I add
echo "<img src='/album/img/".$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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