Re: sub query
От | Christoph Haller |
---|---|
Тема | Re: sub query |
Дата | |
Msg-id | 3F684CF8.64F0359D@rodos.fzk.de обсуждение исходный текст |
Ответ на | sub query ("Martin Kuria" <martinkuria@hotmail.com>) |
Список | pgsql-sql |
The reason why the first one SELECT MAX(e.d), e.host_position FROM (SELECT host_position, COUNT(host_position) as d FROM sss_host GROUP BY host_position) as e GROUP BY e.host_position ORDER BY 1 LIMIT 1; did not work is because the query needs a ORDER BY 1 DESC LIMIT 1 like the second one. Mind the DESCending order. Regards, Christoph > > Thanks Haller, the second one worked thanks a million be blessed > > > > > > > > Hi I have this problem, when I try to run this query: > > > > > > > > SELECT MAX(d), host_position FROM (SELECT host_position, > > > > COUNT(host_position) as d FROM sss_host GROUP BY host_position) as > >e; > > > > > > > > am getting and ERROR: Attribute e.host_position must be GROUPed or > > > used in > > > > an aggregate function. > > > > > > > > Please to advice what could be the problem and how can I rewrite it > >to > > > work > > > > thanks in advance. > > > > > > > As the error message says: e.host_position must be GROUPed > > > > > > so (supposing you want a one row result showing the maximum count) > > > > > > SELECT MAX(e.d), e.host_position FROM (SELECT host_position, > > > COUNT(host_position) as d FROM sss_host GROUP BY host_position) as e > > > GROUP BY e.host_position ORDER BY 1 LIMIT 1; > > > > > > should match your intentions. > > > > >Just thought about another (less complex) way: > > > >SELECT COUNT(host_position), host_position FROM > >sss_host GROUP BY host_position ORDER BY 1 DESC LIMIT 1; > >
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