我在 ORACLE 中的同一个查询中需要 2 个计数列

Question

我正在尝试使用 2 个 count() 函数获取公司已收到和发出的发票的唯一数量。在 invoices table 中有两列引用了同一个公司 ID（一个是发送发票的公司的 ID，另一个是接收发票的公司的 ID）这是我尝试使用的代码：

SELECT K.ID,K.NAME,K.CITY, COUNT(*) AS NUM_OF_INVOICES_SENT, COUNT(*) AS NUM_OF_INVOICES_RECEIVED
FROM COMPANY K LEFT JOIN INVOICE F ON F.COMP_SNEDING = K.ID 
GROUP BY K.NAME,K.ID,K.CITY

这是一个学校项目，所以我一点也不精通sql/oracle

实际数据发票：实际数据公司：给定实际数据的预期结果：

Answer 1

这是一种选择；它不使用 count，而是使用 sum 和 case 表达式。

示例数据：

SQL> with
  2  invoice (id, amount, comp_sending, comp_receiving) as
  3    (select 1, 2000 , 1, 2 from dual union all
  4     select 2, 28250, 3, 2 from dual union all
  5     select 3, 8700 , 4, 1 from dual union all
  6     select 4, 20200, 5, 3 from dual union all
  7     select 5, 21500, 3, 4 from dual
  8    ),
  9  company (id, name, city, state) as
 10    (select 1, 'Microsoft', 'Redmond'  , 'Washington' from dual union all
 11     select 2, 'Ubisoft'  , 'Paris'    , 'France'     from dual union all
 12     select 4, 'Starbucks', 'Seattle'  , 'Washington' from dual union all
 13     select 5, 'Apple'    , 'Cupertino', 'California' from dual union all
 14     select 3, 'Nvidia'   , 'Cupertino', 'California' from dual
 15    )

查询从这里开始：

 16  select c.id, c.name,
 17    sum(case when c.id = i.comp_sending   then 1 else 0 end) cnt_sent,
 18    sum(case when c.id = i.comp_receiving then 1 else 0 end) cnt_received
 19  from company c left join invoice i on c.id in (i.comp_sending, i.comp_receiving)
 20  group by c.id, c.name
 21  order by c.id;

        ID NAME        CNT_SENT CNT_RECEIVED
---------- --------- ---------- ------------
         1 Microsoft          1            1
         2 Ubisoft            0            2
         3 Nvidia             2            1
         4 Starbucks          1            1
         5 Apple              1            0

SQL>

Answer 2

如果将 CASE 表达式中的 0 替换为 NULL，则可以使用 COUNT。所以@Littlefoot 的查询变成了

select c.id, c.name,
  COUNT(case when c.id = i.comp_sending   then 1 else NULL end) cnt_sent,
  COUNT(case when c.id = i.comp_receiving then 1 else NULL end) cnt_received
from company c left join invoice i on c.id in (i.comp_sending, i.comp_receiving)
group by c.id, c.name
order by c.id;

这是有效的，因为 COUNT 只计算那些在被计算的表达式中具有 non-NULL 值的行。

db<>fiddle here

我在 ORACLE 中的同一个查询中需要 2 个计数列

I need 2 count columns in the same query in ORACLE

oracle