Question

我努力让这个SQL查询起作用。我试图在一个查询中获取所需的信息，因为我得到的数据将使用PHP转换为CSV。

我的以下SQL查询是：

SELECT C.customer_ID, C.l_Name AS Surname, 
C.f_Name AS 'First Name', C.travel_Date,
T.tour_Name, 
S.f_Name AS Staff, S.l_Name AS Staff_Surname, 
PA.pay_Method 
FROM Customers AS C 
LEFT JOIN Tour AS T ON C.trip_Taken = T.tour_ID 
LEFT JOIN Staff AS S ON O.staff_ID = S.staff_ID 
LEFT JOIN Payment AS PA ON O.pay_ID = PA.pay_ID 
WHERE C.customer_ID NOT IN (SELECT O.customer_ID 
FROM Customers AS C, Orders AS O 
WHERE C.travel_Date >= '2014-07-08 00:00:00' 
AND C.travel_Date <= '2014-07-08 23:59:59' AND C.customer_ID = O.customer_ID )

我收到此错误#1054 - Unknown column 'O.staff_ID' in 'on clause'并且我不确定如何修复它。任何帮助将不胜感激。运行MySQL 5.0和最新的PHP

Answer 1

你从其他答案得到错误的原因。如果您真的想加入Orders，请在JOIN下添加另一个Orders条款，如下所示

FROM Customers AS C 
LEFT JOIN Tour AS T ON C.trip_Taken = T.tour_ID
LEFT JOIN Orders AS OO ON C.customer_ID = OO.customer_ID <--Included
LEFT JOIN Staff AS S ON OO.staff_ID = S.staff_ID <-- Join with Orders
LEFT JOIN Payment AS PA ON OO.pay_ID = PA.pay_ID

Answer 2

在某些时候你写了

LEFT JOIN Staff AS S ON O.staff_ID = S.staff_ID

但是＆＃34; O＆＃34;在括号范围内定义。在

中更改您的查询

LEFT JOIN Staff AS S ON Orders.staff_ID = S.staff_ID

Answer 3

报告的问题是由于O别名是在子查询中引入的，因此无法在外部联接中使用。

考虑删除子查询，这也解决了初始问题。

SELECT ..
FROM Customers AS C 
JOIN Orders AS O ON C.customer_ID = O.customer_ID -- Join/alias here
LEFT JOIN Tour AS T ON C.trip_Taken = T.tour_ID 
LEFT JOIN Staff AS S ON O.staff_ID = S.staff_ID   -- So it can be used here    
LEFT JOIN Payment AS PA ON O.pay_ID = PA.pay_ID 
-- Promote sub-query conditions
-- (FWIW, the conditions look odd: a customer has only one "travel date"?)
WHERE NOT (C.travel_Date >= '2014-07-08 00:00:00' 
           AND C.travel_Date <= '2014-07-08 23:59:59')

如果客户有 no 订单，则上述查询的行为会有所不同，可以通过根据意图将JOIN更改为LEFT JOIN来解决此问题。

＃1054 - 使用左连接的未知列

3 个答案: