Question

我有一张表，我试图从使用2个左连接的相同数据中获得2个不同的计数。

无论出于何种原因，它都会复制数据并提供不正确的结果，我不确定原因。

这是我到目前为止的查询，我认为它将起作用：

DECLARE @locale INT = '14'

SELECT TOP 50
    E.[DepartmentDesc] AS department,
    COUNT(N.[nomineeQID]) AS totalNominations,
    COUNT(S.[subQID]) AS totalSubmissions,
    COUNT(N.[nomineeQID]) + COUNT(S.[subQID]) AS total
FROM 
    employees AS E
LEFT JOIN 
    submissions AS S ON E.[qid] = S.[subQID] AND S.[statusID] = 3
                        AND S.[locationID] = @locale
LEFT JOIN 
    submissions AS N ON E.[qid] = N.[nomineeQID] AND N.[statusID] = 3
                     AND N.[locationID] = @locale
GROUP BY
    E.[DepartmentDesc]
ORDER BY
    totalNominations DESC

以下是数据的SQL小提琴：http://sqlfiddle.com/#!3/4e6b5/1

结果应该如下，但它提供了偏斜的数字：

总提名应为3
提交的总数应为2
总计应为5

我感觉很接近，但数学不合作！

有什么想法吗？

Answer 1

您正在为每个部门获取笛卡尔积。对查询的最简单修复是使用count(distinct)：

                    COUNT(DISTINCT N.[nomineeQID]) AS totalNominations,
                    COUNT(DISTINCT S.[subQID])     AS totalSubmissions,
                    COUNT(DISTINCT N.[nomineeQID]) + COUNT(DISTINCT S.[subQID]) AS total

更正确的解决方法是在执行join之前在子查询中执行聚合。

编辑：

由于重复问题，请改用SubmissionId：

                    COUNT(DISTINCT N.SubmissionId) AS totalNominations,
                    COUNT(DISTINCT S.SubmissionId)     AS totalSubmissions,
                    COUNT(DISTINCT N.SubmissionId) + COUNT(DISTINCT S.SubmissionId) AS total

Answer 2

试试这个：

DECLARE @locale INT = '14'

select TOP 50 t.department, sum(t.totalSubmissions),
sum(t.totalNominations), sum(t.total)
from 
(SELECT
    E.[DepartmentDesc]    AS department,

    (select COUNT(S.[subQID])
    from submissions AS S
    where E.[qid] = S.[subQID]
    AND S.[statusID] = 3
    AND S.[locationID] = @locale) AS totalSubmissions,

    (select COUNT(N.[nomineeQID])
    from submissions AS N
    where E.[qid] = N.[nomineeQID]
    AND N.[statusID] = 3
    AND N.[locationID] = @locale) AS totalNominations,

    (select COUNT(S.[subQID])
    from submissions AS S
    where E.[qid] = S.[subQID]
    AND S.[statusID] = 3
    AND S.[locationID] = @locale) +
    (select COUNT(N.[nomineeQID])
    from submissions AS N
    where E.[qid] = N.[nomineeQID]
    AND N.[statusID] = 3
    AND N.[locationID] = @locale) AS total  

FROM employees AS E
where exists(
  select 'submission'
  from submissions AS S
  where E.[qid] = S.[subQID]
  AND S.[statusID] = 3
  AND S.[locationID] = @locale
) or
exists(
 select 'nomination'
 from submissions AS N
 where E.[qid] = N.[nomineeQID]
 AND N.[statusID] = 3
 AND N.[locationID] = @locale
 )
) as t
group by t.department
ORDER BY sum(t.totalNominations) DESC

转到SqlFiddle

TSQL加入具有多个计数的表

2 个答案: