Question

我在查询时遇到了一些麻烦，并希望有人可以提供帮助。我尝试过搜索解决方案，但似乎无法找到与我的搜索字词类似的情况。

以下是我要找的内容：

我有一个包含三列的表，需要返回第一列中的值，以及第二列中两个值的计数。一个计数基于参数，另一个计数是总计数。理想情况下，我只想返回第一列中具有相同计数的值。

例如：

Part Number | Make ID
ABC123      | 1<br>
ABC123      | 1<br>
ABC123      | 3<br>
DEF456      | 1<br>
DEF456      | 1

Part Number | Count of Apps Where Make ID = 1| Count of Total Apps
ABC123      |                              2 | 3
DEF456      |                              2 | 2

我到目前为止的查询将返回具有参数值的部件号的应用总数，但是我需要它来返回两个计数：

SELECT apps.part#, 
       COUNT(DISTINCT application#)apps 
FROM   [mytable] AS apps 
       INNER JOIN (SELECT part# 
                   FROM   [mytable] 
                   WHERE  make = '1') AS sz 
               ON sz.part# = apps.part# 
GROUP  BY apps.part# 
ORDER  BY 1

非常感谢任何帮助！

谢谢大家！我收到了几个正确的答案，并选择了第一个给出的答案：

SELECT part#, 
       COUNT(DISTINCT CASE WHEN make = '1' THEN application# END) make1_apps,
       COUNT(DISTINCT application#) total_apps
FROM   [mytable]
GROUP  BY part#
HAVING COUNT(DISTINCT CASE WHEN make = '1' THEN application# END)
     = COUNT(DISTINCT application#)
ORDER  BY part#

再次感谢！

Answer 1

试试这个：

SELECT
    apps.part#
,   COUNT(*)
,   SUM(CASE WHEN make = '1' THEN 1 ELSE 0 END)
FROM [mytable]
GROUP BY apps.part#

此解决方案通过计算总和来替换计数，并为匹配项提供1，为不匹配项提供0。

要过滤掉两个计数不同的记录，请添加HAVING子句：

SELECT
    apps.part#
,   COUNT(*)
,   SUM(CASE WHEN make = '1' THEN 1 ELSE 0 END)
FROM [mytable]
GROUP BY apps.part#
HAVING SUM(CASE WHEN make = '1' THEN 1 ELSE 0 END) = COUNT(*)

Answer 2

SELECT [Part Number],
       SUM(CASE WHEN [Make ID] = 1 THEN 1 ELSE 0 END),
       COUNT(*)
FROM tableName
GROUP BY [Part Number]

SQLFiddle Demo

更新1

SELECT [Part Number],
       SUM(CASE WHEN [Make ID] = 1 THEN 1 ELSE 0 END),
       COUNT(*)
FROM tableName
GROUP BY [Part Number]
HAVING SUM(CASE WHEN [Make ID] = 1 THEN 1 ELSE 0 END) = COUNT(*)

SQLFiddle Demo

Answer 3

另一种更类似于你的方式：

select d.[Part Number], count(d.[Part Number]), o.ones
from
  data d
cross apply
(
  select count(*)
  from data d2
  where d2.[Part Number] = d.[Part Number]
      and d2.[Make ID] = 1
) o (ones)
group by d.[Part Number],  o.ones

SQLFiddle

完成第二次请求添加：

having count(d.[Part Number]) = o.ones

Answer 4

SELECT part#, 
       COUNT(DISTINCT CASE WHEN make = '1' THEN application# END) make1_apps,
       COUNT(DISTINCT application#) total_apps
FROM   [mytable]
GROUP  BY part#
HAVING COUNT(DISTINCT CASE WHEN make = '1' THEN application# END)
     = COUNT(DISTINCT application#)
ORDER  BY part#;

HAVING子句确保两个计数匹配。

使用和不使用参数计算同一列中的记录

4 个答案: