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的情况都在我的结果集中显示一行但是,因为我没有任何值会落在1到4999之间,那行不显示。我仍然希望显示该行,并且只为其列包含0(当然除了第一行)。任何人都可以告诉我如何修改代码来实现这一目标?也许我需要以不同的方式做...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 |
答案 0 :(得分:4)
您可以使用CTE构建查找表,然后使用它对其进行分组而不是Case语句。您需要进行其他三项更改
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
答案 1 :(得分:3)
您可以将描述性文字存储在真实或短暂的表格中。相应地加入,比如;
;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
答案 2 :(得分:2)
您可以添加值,然后插入另一个组(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]
答案 3 :(得分:1)
我从子查询中选择
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
)
我的最佳选择将是 Max(Money * ForReal),所以你得到你的分组,0作为MaxAmount ..