Question

我的Sql查询

 SELECT BOOKING_TIME,
       Contact_No,
       (FName+LName)AS NAME ,
       E_MAIL,

  (SELECT ZM.ZONE_NAME
   FROM Zone_Master ZM
   INNER JOIN BOOKINGS ON ZM.Zone_ID = BOOKINGS.Zone_ID)AS ZONE_NAME,
       City,
       Addr_1,
       Addr_2,
       PIN,

  (SELECT PROJECTS.PROJECT_NAME
   FROM PROJECTS
   INNER JOIN BOOKINGS ON PROJECTS.PROJECT_ID=BOOKINGS.PROJECT_ID)AS PROJECT_NAME
FROM BOOKINGS

Answer 1

由于您的子查询，您收到了该错误：

(SELECT ZM.ZONE_NAME
 FROM Zone_Master ZM
 INNER JOIN BOOKINGS ON ZM.Zone_ID = BOOKINGS.Zone_ID) AS ZONE_NAME

和

(SELECT PROJECTS.PROJECT_NAME
 FROM PROJECTS
 INNER JOIN BOOKINGS ON PROJECTS.PROJECT_ID = BOOKINGS.PROJECT_ID) AS PROJECT_NAME

您将获得多条记录并尝试将它们存储在一个字段中。

Answer 2

以下是您的查询：

SELECT  BOOKING_TIME, Contact_No,(FName+LName)AS NAME, E_MAIL,
        (SELECT ZM.ZONE_NAME
         FROM Zone_Master ZM INNER JOIN
              BOOKINGS
              ON ZM.Zone_ID = BOOKINGS.Zone_ID
       ) AS ZONE_NAME,
       City, Addr_1, Addr_2, PIN,
       (SELECT PROJECTS.PROJECT_NAME
        FROM PROJECTS INNER JOIN
             BOOKINGS
             ON PROJECTS.PROJECT_ID=BOOKINGS.PROJECT_ID
       ) AS PROJECT_NAME
FROM BOOKINGS;

子查询可能返回多行。在select子句的子选择中，您只能返回一个值。我认为有一个简单的解决方法。您可能需要相关的子查询，因此只需从每个子查询中删除BOOKINGS表：

SELECT  BOOKING_TIME, Contact_No, (FName+LName)AS NAME, E_MAIL,
        (SELECT ZM.ZONE_NAME
         FROM Zone_Master ZM
         WHERE ZM.Zone_ID = BOOKINGS.Zone_ID
       ) AS ZONE_NAME,
       City,Addr_1,Addr_2,PIN,
       (SELECT PROJECTS.PROJECT_NAME
        FROM PROJECTS 
        WHERE PROJECTS.PROJECT_ID = BOOKINGS.PROJECT_ID
       )AS PROJECT_NAME
FROM BOOKINGS;

这些现在是“相关子查询”。在这种情况下，他们每个应该最多返回一行。

表达此查询的另一种方法是使用join语法：

SELECT BOOKING_TIME, Contact_No, (FName+LName)AS NAME, E_MAIL,
       ZM.ZONE_NAME,
       City, Addr_1, Addr_2, PIN,
       p.PROJECT_NAME
FROM BOOKINGS b LEFT OUTER JOIN
     Zone_Master zm
     on ZM.Zone_ID = BOOKINGS.Zone_ID LEFT OUTER JOIN
     PROJECTS p
     on p.PROJECT_ID = b.PROJECT_ID

为什么会出现此错误：子查询返回的值超过1

2 个答案: