sql-server - 检查约束在 SQL Server 2016 中不起作用

Question

下面有一个简单的例子

begin tran

CREATE TABLE [dbo].[Filters]
(
    [Id]                    INT NOT NULL IDENTITY,
    [FCode]                 varchar(30) null,
    [FVersion]              varbinary(892) null,

    CONSTRAINT [PK_Filter] PRIMARY KEY CLUSTERED ([Id]),

    CONSTRAINT [CK_Filters_FCode_FVersion]  
        CHECK (([FCode] IS NULL AND [FVersion] IS NULL) 
               OR (LEN([FCode]) > 0 AND DATALENGTH([FVersion]) > 0)), 
)

INSERT INTO [dbo].[Filters] (FCode, FVersion)
VALUES  (NULL, NULL),
        (NULL, 0x6BE348),
        ('ASD', NULL),
        ('ASD', 0x6BE348)

SELECT
    IIF(([FCode] IS NULL AND [FVersion] IS NULL) 
         OR (LEN([FCode]) > 0 AND DATALENGTH([FVersion]) > 0) , 1, 0) AS [check], *
FROM
    [dbo].[filters]

rollback

我预计第二个和第三个插入语句将导致违反约束。但是服务器允许它们。

在使用 select 语句时可以看到第 2 行和第 3 行违反了约束检查的值。查看结果

check   Id  FCode   FVersion
------------------------------
1       1   NULL    NULL
0       2   NULL    0x6BE348
0       3   ASD     NULL
1       4   ASD     0x6BE348

有什么想法吗？

Answer 1

只有当值明确为 false 时，检查约束才会失败。

您希望违反约束的表达式计算为 UNKNOWN。

你可以看到这个

SELECT CASE
         WHEN( ( [FCode] IS NULL
                 AND [FVersion] IS NULL )
                OR ( LEN([FCode]) > 0
                     AND DATALENGTH([FVersion]) > 0 ) ) THEN 'True'
         WHEN NOT ( ( [FCode] IS NULL
                      AND [FVersion] IS NULL )
                     OR ( LEN([FCode]) > 0
                          AND DATALENGTH([FVersion]) > 0 ) ) THEN 'False'
         ELSE 'Unknown'
       END                    AS [check],
       LEN([FCode])           AS LenFCode,
       DATALENGTH([FVersion]) AS DataLengthFVersion,
       *
FROM   [dbo].[filters] 

哪个返回

+---------+----------+--------------------+----+-------+----------+
|  check  | LenFCode | DataLengthFVersion | Id | FCode | FVersion |
+---------+----------+--------------------+----+-------+----------+
| True    | NULL     | NULL               |  1 | NULL  | NULL     |
| Unknown | NULL     | 3                  |  2 | NULL  | 0x6BE348 |
| Unknown | 3        | NULL               |  3 | ASD   | NULL     |
| True    | 3        | 3                  |  4 | ASD   | 0x6BE348 |
+---------+----------+--------------------+----+-------+----------+

它们是 UNKNOWN 的原因是因为 LEN 和 DATALENGTH 在传递 NULL 时都返回 NULL