Re: SQL state: 42P01
От | Tom Lane |
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Тема | Re: SQL state: 42P01 |
Дата | |
Msg-id | 9252.1200356839@sss.pgh.pa.us обсуждение исходный текст |
Ответ на | SQL state: 42P01 ("David Monarchi" <david.e.monarchi@gmail.com>) |
Ответы |
SQL to extract column metadata
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Список | pgsql-novice |
"David Monarchi" <david.e.monarchi@gmail.com> writes: > When I execute the following query, > select A.domain_name_dom, A.word_wdl, A.count_ofi, A.total_quality_ofi, > A.avg_quality_ofi > from zzz_brand2domain_step2_bs2 as A join > (select B.domain_ofi, B.word_ofi, B.count_ofi, B.total_quality_ofi, > B.avg_quality_ofi > from offpagedomainwordintersection_ofi as B > where B.domain_ofi != A.domain_name_dom > order by B.total_quality_ofi desc > limit 1) as C > on A.word_wdl = B.word_ofi; > I receive this error message > ERROR: invalid reference to FROM-clause entry for table "a" > LINE 5: where B.domain_ofi != A.domain_name_dom > ^ > HINT: There is an entry for table "a", but it cannot be referenced from > this part of the query. > I don't understand why I can't reference the respective tables in the > queries. Because a JOIN happens between two *independent* tables. The above isn't well-defined. regards, tom lane
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