Question

SQL查询：

SELECT * 
FROM Account 
WHERE (type <>100000002 ? Id='something': Id=null)

但它显示错误：

'？'
附近的语法不正确

请帮帮我。

Answer 1

这适用于SQL Server。 IIF在SQL Server 2008或更早版本中不可用。

 SELECT * 
    FROM Account 
    WHERE 
    Id=IIF(type<> 100000002,"something",null )

如果您使用的是SQL Server 2008或更早版本，请尝试此操作。

  SELECT * 
    FROM Account 
    WHERE (Id= CASE WHEN type <>100000002 THEN 'something'  ELSE null END)

Answer 2

你可以这样做：

SELECT * 
FROM Account 
where (type <>100000002 and Id='something') or (type =100000002 and id is null)

或

SELECT * 
FROM Account 
where  isnull(id,'_null_')= case when type <>100000002 then 'something' else  isnull(id,'_null_') end

Answer 3

SELECT * 
FROM Account 
WHERE (Id= CASE WHEN type <>100000002 THEN 'something'  ELSE null END)

Answer 4

使用以下内容：

SELECT * 
FROM Account 
WHERE (Id= CASE WHEN type != 100000002 THEN 'something'  ELSE null END)

或

SELECT * 
FROM Account 
WHERE (Id= CASE WHEN type <> 100000002 THEN 'something'  ELSE null END)

Answer 5

SELECT * 
FROM Account 
where (type <>100000002 and Id='something') or (type =100000002)

Answer 6