Question

对于以下查询，我收到此错误：

Msg 156，Level 15，State 1，Line 3：关键字'in'
附近的语法不正确

查询：

SELECT shedule.shedulId, shedule.shedudate, shedule.hometeam, shedule.awayteam
FROM shedule
where in
(Select homeplay.shedulId from homeplay, shedule where shedule.shedulId != homeplay.shedulId);

Answer 1

将评论读到另一个回复，我认为你正在寻找这个：

SELECT h.shedulId, 
       s.shedudate, 
       s.hometeam, 
       s.awayteam
FROM   homeplay h 
LEFT OUTER JOIN  
       shedule s ON 
       h.shedulId = s.shedulId
WHERE  s.shedulId IS NULL

Answer 2

你需要IN操作员面前的东西。

SELECT shedule.shedulId, shedule.shedudate, shedule.hometeam, shedule.awayteam
FROM shedule
where SOMETHING_IS_MISSING_HERE in (Select homeplay.shedulId
                                    from homeplay,shedule
                                    where shedule.shedulId!=homeplay.shedulId)

或者您是否尝试使用IN代替EXISTS？

Answer 3

IN

的示例用法

SELECT p.FirstName, p.LastName, e.JobTitle
FROM Person.Person p
JOIN HumanResources.Employee AS e
    ON p.BusinessEntityID = e.BusinessEntityID
WHERE e.JobTitle IN ('Design Engineer', 'Tool Designer', 'Marketing Assistant');
GO

有关详细信息，请参阅here

Answer 4

正确的陈述是：

SELECT shedule.shedulId, shedule.shedudate, shedule.hometeam, shedule.awayteam
FROM shedule
WHERE shedule.shedulId IN
(SELECT homeplay.shedulId FROM homeplay, shedule WHERE shedule.shedulId != homeplay.shedulId);

Answer 5

请试试这个，它会给你想要的结果

select s.shedulId, s.shedudate, s.hometeam, s.awayteam
from shedule s left join homeplay h on s.shedulId=h.shedulId
where h.shedulId is null

关键字'in'附近的语法不正确

5 个答案: