Vieta tabulā pēc "pts".

[nascar]

Sveiki, lieta tāda ka man iekš DB tabula ir 30 komandas, tālāk es ar šo gribu izvilkt komandas vietu tabulā pēc "pts"

 

Kods:

$team_ranks = mysql_query("SELECT *, pts AS myteamrank, (SELECT COUNT(DISTINCT(pts)) FROM tabula
WHERE pts>myteamrank) AS teamrank FROM tabula WHERE team_saisinajums = '".$_GET['team_saisinajums']."'") or die(mysql_error());
$team_rank = mysql_fetch_array($team_ranks);

 

Un izvadu kā:

<?=$team_rank['teamrank'];?>

 

Kapēc viņš man izvada visām 30 komandām "1" vietu?

Pēc būtības jabūt no 1 > 30, selektojot pēc pts...

 

Ceru ka sapratāt, ko vēlos jums pateikt.

Paldies, Kaspars.

[codez]

... SELECT COUNT(*) ...

Edited October 15, 2011 by codez

[nascar]

Es izdariju tā, bet rāda tagad

 

Reference 'myteamrank' not supported (reference to group function)

 

mysql_query("SELECT COUNT(*) AS myteamrank, (SELECT COUNT(DISTINCT(pts)) FROM tabula
WHERE pts>myteamrank) AS teamrank FROM tabula WHERE team_saisinajums = '".$_GET['team_saisinajums']."' GROUP BY pts") or die(mysql_error());

 

Es pieliku tagad grupu klāt, bet tāpatas nekādas izmaiņas... ES uzreiz saku es ar Grupām nēsmu saskārsies...

[codez]

SELECT *, pts AS myteamrank, (SELECT COUNT(*) FROM tabula WHERE pts>myteamrank) AS teamrank FROM tabula WHERE team_saisinajums='123'

[nascar]

Vienalga izvelk, visiem '1' numuru.

[codez]

Uztaisīju tabulu t2 ar laukiem id,pts. Savadīju pāris vērtība un šis kverijs strādā pareizi:

SELECT *,pts as mypts, (SELECT count(*) as rank FROM t2 WHERE pts>mypts) as myrank FROM t2 WHERE id=300

Edited October 15, 2011 by codez

[nascar]

Ā, nu man arī iet.

 

Paldies, codez.

 

TC.