MySql 连接 UNION 的结果

MySql CONCATENATE result of UNION

我想将查询并集的结果串联起来...我的代码是这样的:

SELECT GROUP_CONCAT(rifDoc), idUser, user, dateDoc
FROM
(
SELECT GROUP_CONCAT(CAR.rifDoc) AS rifDoc, CAR.idUser, CAR.user, CARDETT.dateDoc
FROM car AS CAR, carDett AS CARDETT 
WHERE CAR.id>0 CAR.id=CARDETT.idDoc CARDETT.dateDoc<='2017-01-31' 
GROUP BY idUser, dateDoc

UNION ALL

SELECT GROUP_CONCAT(BK.rifDoc) AS rifDoc, BK.idUser, BK.user, GROUP_CONCAT(BK.inUso) AS inUso, GROUP_CONCAT(BK.inCarico) AS inCarico, BKDETT.dateDoc
FROM bike AS BK, bikeDett AS BKDETT 
WHERE BK.id>0 AND BK.id=BKDETT.idDoc AND BKDETT.dateDoc<='2017-01-31' 
GROUP BY idUser, dateDoc
 )

GROUP BY idUser, dateDoc

但是我有这样的错误:

#1248 - Every derived table must have its own alias

有人有解决办法吗?

您缺少内联视图或子查询的别名,因为错误状态如

SELECT GROUP_CONCAT(rifDoc), idUser, `user`, dateDoc
FROM
(
SELECT GROUP_CONCAT(CAR.rifDoc) AS rifDoc, CAR.idUser, CAR.user, CARDETT.dateDoc
FROM car AS CAR, carDett AS CARDETT 
WHERE CAR.id>0 CAR.id=CARDETT.idDoc CARDETT.dateDoc<='2017-01-31' 
GROUP BY idUser, dateDoc

UNION ALL

SELECT GROUP_CONCAT(BK.rifDoc) AS rifDoc, BK.idUser, BK.user, GROUP_CONCAT(BK.inUso) AS inUso, GROUP_CONCAT(BK.inCarico) AS inCarico, BKDETT.dateDoc
FROM bike AS BK, bikeDett AS BKDETT 
WHERE BK.id>0 AND BK.id=BKDETT.idDoc AND BKDETT.dateDoc<='2017-01-31' 
GROUP BY idUser, dateDoc
 ) XXX     <--- here

每个派生的 table(也称为子查询)确实必须有一个别名。 IE。括号中的每个查询都必须被赋予一个别名(AS 随便什么),可以用来在外部查询的其余部分引用它。 我不确定,但我认为你的代码应该是这样的

SELECT GROUP_CONCAT(rifDoc), idUser, user, dateDoc
FROM
(
SELECT GROUP_CONCAT(CAR.rifDoc) AS rifDoc, CAR.idUser, CAR.user, CARDETT.dateDoc
FROM car AS CAR, carDett AS CARDETT 
WHERE CAR.id>0 CAR.id=CARDETT.idDoc CARDETT.dateDoc<='2017-01-31' 
GROUP BY idUser, dateDoc) As T1

UNION ALL

(SELECT GROUP_CONCAT(BK.rifDoc) AS rifDoc, BK.idUser, BK.user, GROUP_CONCAT(BK.inUso) AS inUso, GROUP_CONCAT(BK.inCarico) AS inCarico, BKDETT.dateDoc
FROM bike AS BK, bikeDett AS BKDETT 
WHERE BK.id>0 AND BK.id=BKDETT.idDoc AND BKDETT.dateDoc<='2017-01-31' 
GROUP BY idUser, dateDoc
 )As T2)

GROUP BY idUser, dateDoc

不确定代码,但解决方案是(As anythin)