SQL：如果不執行所有SELECT Case，我怎樣才能確保所有SELECT Case都被表示？

Question

這是我的代碼

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 

現在我的問題是我希望所有的情況都在我的結果集中顯示一行但是，因為我沒有任何值會落在1到4999之間，那行不會出現。 我仍然希望顯示該行，並且只為其列包含0（當然除了第一行）。 任何人都可以告訴我如何修改代碼來實現這一目標？ 也許我需要以不同的方式做...謝謝！

結果集示例我正在尋找......

  |  [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       |

Answer 1

您可以使用CTE構建查找表，然后使用它來對其進行分組而不是Case語句。 您需要進行其他三項更改

• COUNT將需要為COUNT（t.Money），否則當您預期ZERO時，您將得到1。
• 您可能想要對您的MAX（Money）進行COALESCE，但我不確定它的NULL是什么時候你想要它
• 您實際上不能按MAX（MONEY）訂購，因為它可能為空。 因此，最好使用CTE來控制順序

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

現場演示

Answer 2

您可以將描述性文本存儲在真實或短暫的表格中並相應地加入，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

Answer 3

您可以添加值，然后插入另一個組（無法訪問SSMS進行測試）：

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]

Answer 4

嗯......我要做的：

而不是從表t

我從子查詢中選擇

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
)

我的最佳選擇是Max（Money * ForReal），所以你得到你的分組，0作為MaxAmount ..