如何在具有内部和外部联接的 SUM 函数中使用 Count

Question

感谢您帮助我找到解决这个我无法弄清楚的 SQL 的方法。

在我的 SQL 语句中，我有 1 个内部联接和 1 个外部联接。其中一个求和需要计算从内部连接返回的行数

Table 1             

Location |Date       |Column3 |Column4
1        |11/22/2019 |23      |3
2        |11/22/2019 |23      |4
3        |11/22/2019 |12      |5
4        |11/22/2019 |54      |6
5        |11/22/2019 |65      |7
6        |11/22/2019 |23      |8
7        |11/22/2019 |5       |9

Table 2         

Location |Date       |Column3 |Column4 |Column5
1        |11/22/2019 |1       |8.5     |06
1        |11/22/2019 |2       |8.5     |06
1        |11/22/2019 |3       |8       |06
1        |11/22/2019 |4       |8.5     |06
2        |11/22/2019 |5       |7.5     |06 
2        |11/22/2019 |5       |8.5     |06
2        |11/22/2019 |6       |8.5     |06

Table 3         
Location |Date       |Column3 |Column4 |Column5 
1        |11/22/2019 |3       |3       |1
2        |11/22/2019 |4       |4       |0
3        |11/22/2019 |6       |5       |1
4        |11/22/2019 |25      |6       |0
5        |11/22/2019 |65      |7       |1
6        |11/22/2019 |65      |8       |0
7        |11/22/2019 |78      |9       |1

我的错误陈述，需要在第 5 行计数

SELECT 
T3.Location AS Location, 
T3.Date AS [Date],
SUM(T1.Column3- (T2.Column3+ T2.Column4)) as Result_Numerator,
SUM(T1.Column4 - COUNT(T2.Column2)) AS Result_Denominator,
SUM(T3.Column3 - T3.Column4) as Plan_Numerator,
SUM(T3.Column4 - T3.Column3) as Plan_Denominator,
'Element Name' as Element

FROM Table1 T1 
INNER JOIN 
Table2 T2 ON T1.Date= T2.Date AND T1.Location = T2.Location
RIGHT OUTER JOIN
Table3 T3 ON T1.Date = T3.DATE and T3.Location = P.Location

WHERE T1.Location = '1' AND (T3.Date>({ts '2019-11-01 00:00:00'})) 
AND T3.Column5 = '0' AND T2.Column3>0 AND T2.Column5 ('05','06')

GROUP BY T3.Location, 
T1.CNY_CD,
T3.PLN_DATE,

我查看了一些类似的问题，他们建议先使用 union all 来获取计数，但我不完全理解如何使用 T1 和 T2 之间的内部连接来实现这一点。

非常感谢您帮助我 D

Answer 1

RIGHT JOIN 令人困惑。 请改用 LEFT JOIN。

Table2 可能还需要 LEFT JOIN。
但是，Table2 上的标准都应该在 ON 中，而不是在 WHERE 子句中。或者它的行为就像一个 INNER JOIN。

...
FROM Table3 T3
LEFT JOIN Table2 T2 
  ON T2.Date= T3.Date 
 AND T2.Location = T3.Location
 AND T2.Column3 > 0 
 AND T2.Column5 IN ('05','06')
LEFT JOIN Table1 T1 
  ON T1.Date = T3.Date
 AND T1.Location = T3.Location
WHERE T3.Location = '1' 
  AND T3.Date >= CAST('2019-11-01' AS DATE)
  AND T3.Column5 = '0' 
GROUP BY T3.Location, T3.Date

Answer 2

select inner_sum = (Select SUM(...) as inner_sum from Table 1
inner join Table2 on Table2.Date = Table1.Date and Table2.Location = Table1.location), outer_sum = SUM(....)
... the rest of your statement 

阅读更多：SQL Subquery

看来可以直接将子查询的结果赋值给列名，功能相当强大！但是子查询很慢，我相信使用联合有更好的方法。

Select SUM(...) as inner_sum from Table1 
inner join Table2 on Table2.Date = Table1.Date and Table2.location = Table1.location
UNION
Select SUM(...) as outer join
....

Answer 3

您没有在请求中显示预期结果，也没有告诉我们表的键是什么。我假设以下键：

• table1（位置、日期）
• table2（无）
• table3（位置、日期）

似乎（但我可能错了）您希望每个位置和日期都有一个结果行。这意味着您必须聚合 table2 数据。

我不知道您的查询中哪些表数据是必需的，哪些不是。看起来您想从 table3 中进行选择，它可能在 table1 和 table2 中有匹配项，也可能没有匹配项。

许多假设，因为您的请求缺少所有这些信息。下次请更准确地描述您的表格、它们的关系以及您想要的结果。

with t2 as
(
  select location, date, sum(column3) as sum3, sum(column4) as sum4, count(*) as cnt
  from table2
  where column3 > 0 and column5 in ('05', '06')
  group by location, date
)
select
  t3.location, 
  t3.date,
  t1.column3- (t2.sum3 + t2.sum4) as result_numerator,
  t1.column4 - t2.cnt as result_denominator,
  t3.column3 - t3.column4 as plan_numerator,
  t3.column4 - t3.column3 as plan_denominator
from table3 t3
left join table1 t1 on t1.location = t3.location and t1.date = t3.date
left join t2 on t2.location = t3.location and t2.date = t3.date
where t3.location = 1 and t3.date >= '2019-11-02' and t3.column5 = 0
order by t3.location, t3.date;

根据所有这些猜测，您可能需要调整此查询，但您明白了。