SQLIte中嵌套内部联接的问题 [英] problem with nested inner joins in SQLIte

问题描述

以下sql语句无法在SQLite中运行:

 选择*来自AA.optionalfkB = B.pk上的左联接(B.fkC上的B内部联接C = C.pk) 

我得到sqlException未知列B.pk"

根据文档@ http://www.sqlite.org/lang_select.html 这应该工作，并且将在所有其他sql实现中工作.我在做错什么吗?

解决方案

它不起作用，因为外部"查询不知道B是什么.

 选择*来自A左联接(B.fkC = C.pk上的B内部联接C)B. A.optionalfkB = B.pk 

(B.fkC = C.pk上的B内部联接C)很奇怪，没有任何 select ，但是规范确实说这是有效的.

The sql statement below will not run in SQLite:

select * 
from A
left join (B inner join C on B.fkC = C.pk) on A.optionalfkB = B.pk

I get a sqlException "unknown column B.pk"

According to the documentation @ http://www.sqlite.org/lang_select.html this should work, and it will work in all other sql implementations. Am I doing something wrong?

解决方案

It doesn't work because the "outer" query doesn't know what B is.

select * 
from A
left join (B inner join C on B.fkC = C.pk) B on A.optionalfkB = B.pk

The (B inner join C on B.fkC = C.pk) is weird without any select, but the specification does say that it is valid.

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