You have a problem with "...
" is reserved word, put \ before it...
try this:
$sql = "select id, siteurl, sitename, from sitesats where validated =
\"$validated\"";
regards,
Augusto
On Sun, 10 Jun 2001, Vladimir Kravtsov wrote:
> <html>
> <head>
> <body>
> <?php require('../common.php');
> connectdb();
> mysql_select_db(xtopsites);
> $validated=1;
> $sql = "select id, siteurl, sitename, from sitesats where validated = "$validated"";
> // Line 8
>
> $r = mysql_query($sql); //You can ignore this line and on since its not relevant to
>my problem.
> $numrows = mysql_num_rows($r);
> echo (mysql_error());
> $row = mysql_fetch_array($r);
> ?>
> </body>
> </html>
>
> When I run this I get: Parse error: parse error in
>c:\inetpub\wwwroot/php/X-TopSites/admin/admin.php on line 8
> I want it to select the id, siteurl, sitename (in a table called sitesats) of any
>site which is not validated (I made it so validated=1 and not validated=2)
>
> What's wrong here?
> Thanks alot :)
>
>
>
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