I tried what you said but i get an eval error:
Parse error: parse error, unexpected '=' in /index.php(135) : eval()'d
code on line 1
;(
----- Original Message -----
From: "Jay Blanchard" <[EMAIL PROTECTED]>
To: "WebMaster. Radio ECCA" <[EMAIL PROTECTED]>;
<[EMAIL PROTECTED]>
Sent: Thursday, August 19, 2004 1:38 PM
Subject: RE: [PHP] Links with parameters in DB
[snip]
It doesn´t work, I have other fields in the DB apart from the field 'Link',
so if I use
$link = eval($result);
I get a parse error. Apart from that, I have to write the name of the field
(link), if not the server won´t know the field I´m refering to.
[/snip]
Then did you eval that? I meant for you to use a proper rendering of the
eval statement by itself, i.e.
$sql = "SELECT * FROM table ";
$result = mysql_query($sql, $connection);
while($row = mysql_fetch_array($result)){
$link = eval($row['Link']);
echo $link;
}
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