问题描述

下图是 Microsoft SQL Server 2008 R2 系统视图的一部分.从图中我们可以看出，sys.partitions 和sys.allocation_units 的关系取决于sys.allocation_units.type 的值.所以为了把它们结合在一起，我会写一些类似的东西:

The following image is a part of Microsoft SQL Server 2008 R2 System Views. From the image we can see that the relationship between sys.partitions and sys.allocation_units depends on the value of sys.allocation_units.type. So to join them together I would write something similar to this:

SELECT  *
FROM    sys.indexes i
        JOIN sys.partitions p
            ON i.index_id = p.index_id 
        JOIN sys.allocation_units a
            ON CASE
               WHEN a.type IN (1, 3)
                   THEN a.container_id = p.hobt_id 
               WHEN a.type IN (2)
                   THEN a.container_id = p.partition_id
               END 

但是上面的代码给出了一个语法错误.我猜这是因为 CASE 语句.谁能帮忙解释一下?

But the upper code gives a syntax error. I guess that's because of the CASE statement. Can anyone help to explain a little?

添加错误信息:

消息 102，级别 15，状态 1，第 6 行 '=' 附近的语法不正确.

Msg 102, Level 15, State 1, Line 6 Incorrect syntax near '='.

推荐答案

CASE 表达式从子句的 THEN 部分返回一个值.你可以这样使用它:

A CASE expression returns a value from the THEN portion of the clause. You could use it thusly:

SELECT  * 
FROM    sys.indexes i 
    JOIN sys.partitions p 
        ON i.index_id = p.index_id  
    JOIN sys.allocation_units a 
        ON CASE 
           WHEN a.type IN (1, 3) AND a.container_id = p.hobt_id THEN 1
           WHEN a.type IN (2) AND a.container_id = p.partition_id THEN 1
           ELSE 0
           END = 1

请注意，您需要对返回值进行处理，例如将其与 1 进行比较.您的语句试图返回赋值或相等性测试的值，这在 CASE/THEN 子句的上下文中都没有意义.(如果 BOOLEAN 是一种数据类型，那么相等性测试就有意义.)

Note that you need to do something with the returned value, e.g. compare it to 1. Your statement attempted to return the value of an assignment or test for equality, neither of which make sense in the context of a CASE/THEN clause. (If BOOLEAN was a datatype then the test for equality would make sense.)

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