Lidt avanceret

den finder ikke et id i den første tabel...

Jeg fatter ikke en bjælde af hvad du vil have sql'en skal gøre...

Prøv at smide en description af de to tabeller, så man kan se, hvilke felter de indeholder og fortæl, hvordan de er relaterede.

Eks:
Tabel1:
Id: Brugernavn:

Table2:
Id: Reklamer:

Så kan du godt se i scriptet, at den finder et id i Tabel2. Men der står intet brugernavn. Og det er et brugernavn jeg vil have frem. Så t_owner i tabel2 skal faktisk erstattes af Brugernavn i Tabel1. t_owner i tabel2 er id'et som er i tabel1. Hvis du forstår nu..

SELECT t1.t_owner, t1.COUNT(*) AS hits, t2.username FROM tb_things AS t1, tabel2navn AS t2 WHERE t1.id = t2.id AND t1.t_info = 'Reklame' GROUP BY t1.t_owner ORDER BY hits DESC LIMIT 3

Dette virker ikke:
<?
include("chat_mysql_con.php");
$result= mysql_query("SELECT t1.t_owner, t1.COUNT(*) AS hits, t2.username FROM tb_things AS t1, tb_users AS t2 WHERE t1.id = t2.id AND t1.t_info = 'Reklame' GROUP BY t1.t_owner ORDER BY hits DESC LIMIT 10");
while ($data = mysql_fetch_row($result))
echo "Brugernavn: ".$data[0]." - Online tid: ".$data[1]."<br>";
?>

Warning: Supplied argument is not a valid MySQL result resource in /web/www.skyskraber.dk/www/Chat/hits.php on line 5

$result= mysql_query("SELECT t2.username, COUNT(t1.*) AS hits FROM tb_things AS t1, tb_users AS t2 WHERE t1.t_owner= t2.id AND t1.t_info = 'Reklame' GROUP BY t2.username ORDER BY hits DESC LIMIT 10");

Rettelse, jvf. proaccess:
$result= mysql_query("SELECT t1.t_owner, COUNT(t1.*) AS hits, t2.username FROM tb_things AS t1, tb_users AS t2 WHERE t1.id = t2.id AND t1.t_info = 'Reklame' GROUP BY t1.t_owner ORDER BY hits DESC LIMIT 10");

Her er koden:
<b>Flest reklamer:</b><br>
<?
include("chat_mysql_con.php");
$result= mysql_query("$result= mysql_query("SELECT t1.t_owner, COUNT(t1.*) AS hits, t2.username FROM tb_things AS t1, tb_users AS t2 WHERE t1.id = t2.id AND t1.t_info = 'Reklame' GROUP BY t1.t_owner ORDER BY hits DESC LIMIT 10");");
while ($data = mysql_fetch_row($result))
echo "Brugernavn: ".$data[0]." - Online tid: ".$data[1]."<br>";
?>
<br>
<b>Online mest tid:</b><br>
<?
include("chat_mysql_con.php");
$result= mysql_query("SELECT username, time_all FROM tb_users ORDER BY time_all DESC LIMIT 10");
while ($data = mysql_fetch_row($result))
echo "Brugernavn: ".$data[0]." - Online tid: ".$data[1]."<br>";
?>
<br>
<b>Flest point:</b><br>
<?
include("chat_mysql_con.php");
$result= mysql_query("SELECT username, point FROM tb_users ORDER BY point DESC LIMIT 10");
while ($data = mysql_fetch_row($result))
echo "Brugernavn: ".$data[0]." - Point: ".$data[1]."<br>";
?>
<br>
<b>Flest profil besøg:</b><br>
<?
include("chat_mysql_con.php");
$result= mysql_query("SELECT username, pro_looks FROM tb_users ORDER BY pro_looks DESC LIMIT 10");
while ($data = mysql_fetch_row($result))
echo "Brugernavn: ".$data[0]." - Profil: ".$data[1]."<br>";
?>

Parse error: parse error in /web/www.skyskraber.dk/www/Chat/hits.php on line 4

Rettede selv, her er den nye fejl så. Jeg har ændret i koden:

<b>Flest reklamer:</b><br>
<?
include("chat_mysql_con.php");
$result = mysql_query("SELECT t1.t_owner, COUNT(t1.*) AS hits, t2.username FROM tb_things AS t1, tb_users AS t2 WHERE t1.id = t2.id AND t1.t_info = 'Reklame' GROUP BY t1.t_owner ORDER BY hits DESC LIMIT 10");
while ($data = mysql_fetch_row($result))
echo "Brugernavn: ".$data[0]." - Online tid: ".$data[1]."<br>";
?>
<br>
<b>Online mest tid:</b><br>
<?
include("chat_mysql_con.php");
$result= mysql_query("SELECT username, time_all FROM tb_users ORDER BY time_all DESC LIMIT 10");
while ($data = mysql_fetch_row($result))
echo "Brugernavn: ".$data[0]." - Online tid: ".$data[1]."<br>";
?>
<br>
<b>Flest point:</b><br>
<?
include("chat_mysql_con.php");
$result= mysql_query("SELECT username, point FROM tb_users ORDER BY point DESC LIMIT 10");
while ($data = mysql_fetch_row($result))
echo "Brugernavn: ".$data[0]." - Point: ".$data[1]."<br>";
?>
<br>
<b>Flest profil besøg:</b><br>
<?
include("chat_mysql_con.php");
$result= mysql_query("SELECT username, pro_looks FROM tb_users ORDER BY pro_looks DESC LIMIT 10");
while ($data = mysql_fetch_row($result))
echo "Brugernavn: ".$data[0]." - Profil: ".$data[1]."<br>";
?>


Warning: Supplied argument is not a valid MySQL result resource in /web/www.skyskraber.dk/www/Chat/hits.php on line 5

$result = mysql_query("SELECT t1.t_owner, COUNT(t1.*) AS hits, t2.username FROM tb_things AS t1, tb_users AS t2 WHERE t1.id = t2.id AND t1.t_info = 'Reklame' GROUP BY t1.t_owner ORDER BY hits DESC LIMIT 10") or die(mysql_error());

You have an error in your SQL syntax near '*) AS hits, t2.username FROM tb_things AS t1, tb_users AS t2 WHERE t1.id = t2.id' at line 1

...gider ikke mere...

okay :-) øv øv øv