4

这是我的代码

SELECT
  CASE
       WHEN Money >= 20000 THEN '$ 20,000 + '
       WHEN Money BETWEEN 10000 AND 19999 THEN '$ 10,000 - $ 19,999'
       WHEN Money BETWEEN  5000 AND  9999 THEN '$  5,000 - $  9,999'
       WHEN Money BETWEEN     1 AND  4999 THEN '$      1 - $  4,999'
       ELSE '$      0'
   END AS [MONEY],
       COUNT(*) AS [#],
       MAX(Money) AS [MAX]
  FROM MyTable
 WHERE MoneyType = 'Type A'
 GROUP BY 
  CASE
       WHEN Money >= 20000 THEN '$ 20,000 + '
       WHEN Money BETWEEN 10000 AND 19999 THEN '$ 10,000 - $ 19,999'
       WHEN Money BETWEEN  5000 AND  9999 THEN '$  5,000 - $  9,999'
       WHEN Money BETWEEN     1 AND  4999 THEN '$      1 - $  4,999'
       ELSE '$      0'
   END
 ORDER BY MAX DESC

现在我的问题是我希望所有案例都在我的结果集中显示一行，但是，因为我没有任何介于 1 和 4999 之间的值，所以该行不会显示。我仍然希望该行显示出来，并且只包含 0 的列（当然第一列除外）。任何人都可以告诉我如何修改代码来完成这个？也许我需要用不同的方式来做......谢谢！

我正在寻找的结果集示例...

  |  [MONEY]              |   [#]   |    [MAX]    |
  |  $ 20,000+            |   2     |    30,000   |
  |  $ 10,000 - $ 19,999  |   8     |    19,000   |
  |  $  5,000 - $  9,999  |   4     |     8,000   |
  |  $      1 - $  4,999  |   0     |     0       |     <-- Row currently doesn't show
  |  $      0             |   12    |     0       |
4

4 回答 4

4

您可以使用 CTE 构建一个查找表，然后使用该表而不是 Case 语句对其进行分组。您需要进行其他三项更改

  • COUNT 必须是 COUNT(t.Money)，否则当您期望零时您将得到 1。
  • 您可能想合并您的 MAX(Money) 但我不确定当它为 NULL 时您希望它是什么
  • 您实际上不能按 MAX(MONEY) 订购，因为它可能为空。所以最好也使用 CTE 来控制订单


WITH Ranges AS
( SELECT 1 id , '$ 20,000 +'  description
  UNION SELECT 2 , '$ 10,000 - $19,999'
  UNION SELECT 3, '$  5,000 - $ 9,999'
  UNION SELECT 4, '$      1 - $ 4,999'
  UNION SELECT 5, '$      0')


SELECT
       r.Description as money,
       COUNT(t.Money) AS [#],
       MAX(Money) AS [MAX]
  FROM 
       Ranges r
       LEFT JOIN
        MyTable t
       ON r.ID =   CASE
                    WHEN Money >= 20000 THEN 1
                    WHEN Money BETWEEN 10000 AND 19999 THEN 2
                    WHEN Money BETWEEN  5000 AND  9999 THEN 3
                    WHEN Money BETWEEN     1 AND  4999 THEN 4
                    ELSE 5
                 END 
        AND  MoneyType = 'Type A' 
 GROUP BY 
  r.id,
  r.Description
 ORDER BY r.id asc

现场演示

于 2012-05-17T15:21:05.700 回答
3

您可以将描述性文本存储在真实或临时表中并相应地加入，例如；

;with ranges(min,max,caption) as (
    select 10000, 19999, '$ 10,000 - $19,999' union
    select 5000, 9999,   '$  5,000 - $ 9,999' union
    select 1, 4999,      '$      1 - $ 4,999'
)
select
    isnull(r.caption, 'no description'),
    count(m.money) as [#],
    isnull(max(m.money), 0)
from
    mytable m
full outer join 
    ranges r on m.money between r.min and r.max
group by
    r.caption
于 2012-05-17T15:22:24.490 回答
2

您可以添加这些值，然后插入一个附加组（无法访问 SSMS 进行测试）：

SELECT [MONEY], SUM([#]) AS [#], MAX([MAX]) AS [MAX] FROM
(SELECT
  CASE
       WHEN Money >= 20000 THEN '$ 20,000 + '
       WHEN Money BETWEEN 10000 AND 19999 THEN '$ 10,000 - $19,999'
       WHEN Money BETWEEN  5000 AND  9999 THEN '$  5,000 - $ 9,999'
       WHEN Money BETWEEN     1 AND  4999 THEN '$      1 - $ 4,999'
       ELSE '$      0'
   END AS [MONEY],
       COUNT(*) AS [#],
       MAX(Money) AS [MAX]
  FROM MyTable
 WHERE MoneyType = 'Type A'
 GROUP BY 
  CASE
       WHEN Money >= 20000 THEN '$ 20,000 + '
       WHEN Money BETWEEN 10000 AND 19999 THEN '$ 10,000 - $19,999'
       WHEN Money BETWEEN  5000 AND  9999 THEN '$  5,000 - $ 9,999'
       WHEN Money BETWEEN     1 AND  4999 THEN '$      1 - $ 4,999'
       ELSE '$      0'
   END
 UNION ALL
 SELECT '$      0' AS [MONEY], 0 AS [#], 0 AS [MAX]
 UNION ALL
 SELECT '$      1 - $ 4,999' AS [MONEY], 0 AS [#], 0 AS [MAX]
 UNION ALL
 SELECT '$  5,000 - $ 9,999' AS [MONEY], 0 AS [#], 0 AS [MAX]
 UNION ALL
 SELECT '$ 10,000 - $19,999' AS [MONEY], 0 AS [#], 0 AS [MAX]
 UNION ALL
 SELECT '$ 20,000 + ' AS [MONEY], 0 AS [#], 0 AS [MAX]
 ) SUB
GROUP BY [MONEY]
于 2012-05-17T15:14:58.113 回答
1

嗯......我会做什么：

而不是从表 t

我会从子查询中选择

From 
(
 Select Money, 1 as ForReal from Table t
 Union
 Select 1, 0
 Union
 Select 5000, 0
 Union 
 Select 10000, 0
 Union
 Select 20001, 0
)

我的首选是 Max (Money*ForReal)，所以你得到你的分组，0 作为 MaxAmount..

于 2012-05-17T15:21:05.013 回答