从非序列号中查找缺失的号码

Question

我有一个包含 5 列 (

A1,A2,A3,A4,A5

) 的数据库，其中存储 5 个数字。

这 5 个数字是

"1,2,3,4,5"

。

A1     A2     A3     A4     A5
-------------------------------
2      4      5      Null  Null

我想得到丢失的数字“1”和“3”。 如何从5个号码中找出缺失的号码？

Answer 1

Select Replace(Replace(Replace(Replace(
Replace('12345',(Cast(Coalesce(A5,0) as varchar(1))),''),
(Cast(Coalesce(A4,0) as varchar(1))),''),
(Cast(Coalesce(A3,0) as varchar(1))),''),
(Cast(Coalesce(A2,0) as varchar(1))),''),
(Cast(Coalesce(A1,0) as varchar(1))),'') from Table1

Sql Fiddle 演示

Answer 2

你可以这样做

WITH sequence AS
(
  SELECT 1 n UNION ALL
  SELECT n + 1 FROM sequence WHERE n < 5
)
SELECT n
  FROM sequence s LEFT JOIN table1 t
    ON s.n IN (t.a1, t.a2, t.a3, t.a4, t.a5)
 WHERE t.a1 IS NULL

输出：

|尼 |
|---|
| 1 |
| 3 |

这里是SQLFiddle演示

Answer 3

一个优雅的解决方案可以扩展到您问题中提供的 5 个值之外，它使用了数字表（如链接所示，数字表对于多种任务来说是无价的）。

这是数字表的设置，在本例中限制为 256 个数字，因为您在问题中使用的范围很小。

/* 
    create a numbers table with numbers from 0 to 
    255 (the range of the tinyint data type).
*/
DROP TABLE IF EXISTS [#numbers];
GO

CREATE TABLE [#numbers]
(
    [n] tinyint NOT NULL
);

WITH [numbers]
AS
(
    SELECT [v].[n]
    FROM
    (
        VALUES (0), (1), (2), (3), (4), (5), (6), (7), (8), (9)
    ) [v]([n])
)
INSERT INTO [#numbers]
(
    [n]
)
SELECT TOP(255)
    [number] = 
                  ([n1].[n] * 100)
                + ([n2].[n] * 10) 
                + ([n3].[n] * 1)
FROM 
    [numbers] [n1]
    CROSS JOIN [numbers] [n2]
    CROSS JOIN [numbers] [n3]
ORDER BY
      ([n1].[n] * 100)
    + ([n2].[n] * 10) 
    + ([n3].[n] * 1);

接下来，我们将创建您在问题中提到的表：

DROP TABLE IF EXISTS [#rows];
CREATE TABLE [#rows]
(
      [a1]  tinyint  NULL
    , [a2]  tinyint  NULL
    , [a3]  tinyint  NULL
    , [a4]  tinyint  NULL
    , [a5]  tinyint  NULL
);

在这里，我们用您的值填充该行：

INSERT INTO [#rows] ([a1], [a2], [a3])
VALUES (2, 4, 5);

[#rows]

表的内容：

a1	a2	a3	a4	a5
2	4	5	空	空

我将使用 UNPIVOT 子句 将列转换为行。为了清楚起见，这就是

UNPIVOT

本身的样子：

SELECT 
    [upv].[value]
FROM
(
    SELECT *
    FROM [#rows]
) [src]
UNPIVOT
(
    [value] FOR [column] IN ([a1], [a2], [a3], [a4], [a5])
) [upv]
ORDER BY
    [upv].[value];

UNPIVOT

查询的中间结果：

价值
2
4
5

如您所见，我们缺少值

1

和

3

。在这里，我们通过将

UNPIVOT

的结果与 EXCEPT T-SQL 运算符 与

[#numbers]

表相结合来显示缺失值：

SELECT
    [missing_number] = [n].[n]
FROM
    [#numbers] [n]
WHERE 
    [n].[n]     > 0
    AND [n].[n] <= 5
EXCEPT
SELECT 
    [upv].[value]
FROM
(
    SELECT *
    FROM [#rows]
) [src]
UNPIVOT
(
    [value] FOR [column] IN ([a1], [a2], [a3], [a4], [a5])
) [upv];

EXCEPT

运算符显示了运算符左侧的不同结果，而运算符右侧的结果中不存在这些结果。结果是：

缺少号码
1
3

Answer 4

您需要一个从 1 到（在本例中）5 的整数表：

DECLARE @ints table (n int);
INSERT @ints VALUES (1), (2), (3), (4), (5);

其次，我们将表行中的数字放入一个可比较的集合中：

SELECT x INTO #all FROM (
    SELECT A1 as x FROM myTable WHERE ID = myRow
    UNION ALL
    SELECT A2 as x FROM myTable WHERE ID = myRow
    UNION ALL
    SELECT A3 as x FROM myTable WHERE ID = myRow
    UNION ALL
    SELECT A4 as x FROM myTable WHERE ID = myRow
    UNION ALL
    SELECT A5 as x FROM myTable WHERE ID = myRow
    ) y

那么你就可以得出答案：

SELECT @ints.n
FROM @ints left join #all on @ints.n = #all.x
WHERE #all.x is null
ORDER BY 1