On Thu, 19 Aug 2004 14:01:45 +0100, WebMaster. Radio ECCA
<[EMAIL PROTECTED]> wrote:
> 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 $link;
> }
>
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>
try this
eval("\$link = \"$row[Link]\";");
echo $link;
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