Question

当我尝试加入3张桌子时遇到了问题。

以下是我使用的mysql语法：

SELECT
FROM carecell
LEFT JOIN staff ON cc.id_pks = staff.id_emp
LEFT JOIN staff ON cc.id_wpks = staff.id_emp
INNER JOIN member ON member.id_member = staff.id_member

请帮帮我..我该怎么办才能修复语法？

Answer 1

SQL引擎无法区分staff子句中的两个from表。你需要给他们不同的名字。像这样：

FROM carecell cc LEFT JOIN
     staff s1
     ON cc.id_pks = s1.id_emp LEFT JOIN
     staff s2
     ON cc.id_wpks = s2 .id_emp INNER JOIN
     member m
     ON m.id_member = s2.id_member

Answer 2

如果多次加入同一个表，则每次都需要为该表提供不同的别名，以便数据库引擎可以在它们之间有所不同

SELECT *
FROM carecell as cc
LEFT JOIN staff as s1 ON cc.id_pks = s1.id_emp
LEFT JOIN staff as s2 ON cc.id_wpks = s2.id_emp
INNER JOIN member as m ON m.id_member = s1.id_member

您已经使用了别名，因为您提供了carecell别名cc。在您的查询的最后一行，我加入了s1，但您必须决定采取哪一项 - s1或s2

Answer 3

SELECT *
FROM carecell
LEFT JOIN staff t1 ON cc.id_pks = t1.id_emp
LEFT JOIN staff t2 ON cc.id_wpks = t2.id_emp
INNER JOIN member ON member.id_member = t1.id_member

＃1066 - 不是唯一的表/别名：＆＃39; tb_X＆＃39;

3 个答案: