[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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