2

所以我有两个支付表，我想在 Oracle SQL DB 中进行比较。我想使用位置和发票以及总付款来比较总付款。它比这更复杂，但基本上它是：

select
  tbl1.location,
  tbl1.invoice,
  Sum(tbl1.payments),
  Sum(tbl2.payments)    
From 
  tbl1    
  left outer join tbl2 on 
    tbl1.location = tbl2.location 
    and tbl1.invoice = tbl2.invoice    
group by 
  (tbl1.location,tbl1.invoice)

我想要左外连接，因为除了比较付款金额之外，我还想查看 tbl1 中可能不存在于 tbl2 中的所有订单。

问题是两个表中的每个订单（位置和发票）都有多个记录（不一定相同数量的记录，即 tbl1 中的 2 到 tbl2 中的 1 或反之亦然），但每个订单的总付款（位置和发票）应该匹配。因此，只需进行直接连接就可以得到笛卡尔积。

所以我想我可以做两个查询，首先按商店和发票汇总总付款，然后对这些结果进行连接，因为在汇总结果中，每个订单（商店和发票）只有一条记录。但我不知道该怎么做。我已经尝试了几个子查询，但似乎无法撼动笛卡尔积。我希望能够在一个查询中执行此操作，而不是创建表并加入这些表，因为这将持续进行。

提前感谢您的帮助。

4

3 回答 3

1

您可以使用该With语句创建两个查询，然后按照您的说法加入。我只会放sintaxe，如果您需要更多帮助，请询问。那是因为您没有提供有关表格的完整详细信息。所以我只是猜测我的答案。

WITH tmpTableA as ( 
        select
          tbl1.location,
          tbl1.invoice,
          Sum(tbl1.payments) totalTblA
        From 
          tbl1
        group by 
          tbl1.location,
          tbl1.invoice
         ),
   tmpTableB as ( 
        select
          tbl2.location,
          tbl2.invoice,
          Sum(tbl2.payments) totalTblB
        From 
          tbl2
        group by 
          tbl2.location,
          tbl2.invoice
         )
Select tmpTableA.location,  tmpTableA.invoice, tmpTableA.totalTblA,
       tmpTableB.location,  tmpTableB.invoice, tmpTableB.totalTblB
  from tmpTableA, tmpTableB
 where tmpTableA.location = tmpTableB.location (+)
   and tmpTableA.invoice = tmpTableB.invoice (+)

(+)运算符是Oracle数据库的运算left join符（当然，如果您愿意，可以使用LEFT JOIN语句）

于 2013-12-12T19:01:21.897 回答
0

对不起，我的第一个答案是错误的。感谢您提供 sqlfiddle，MT0。

我错过的一点是你需要先总结每张桌子上的付款，所以每张桌子上只剩下一行，然后加入它们。这就是 MT0 在他的声明中所做的。

如果您想要一个看起来更“对称”的解决方案，请尝试：

select A.location, A.invoice, B.total sum1, C.total sum2
from (select distinct location, invoice from tbl1) A
left outer join (select location, invoice, sum(payments) as total from tbl1 group by location, invoice) B on A.location=B.location and A.invoice=B.invoice
left outer join (select location, invoice, sum(payments) as total from tbl2 group by location, invoice) C on A.location=C.location and A.invoice=C.invoice

这导致

LOCATION    INVOICE SUM1    SUM2
a           2       3       2
a           1       5       3
b           1       1       5
b           2       1       (null)
于 2013-12-12T19:03:44.237 回答
0

另外两个选项：

SQL小提琴

Oracle 11g R2 模式设置：

CREATE TABLE tbl1 ( id, location, invoice, payments ) AS 
          SELECT  1, 'a', 1, 1 FROM DUAL
UNION ALL SELECT  2, 'a', 1, 1 FROM DUAL
UNION ALL SELECT  3, 'a', 1, 1 FROM DUAL
UNION ALL SELECT  4, 'a', 1, 1 FROM DUAL
UNION ALL SELECT  5, 'a', 1, 1 FROM DUAL
UNION ALL SELECT  6, 'a', 2, 1 FROM DUAL
UNION ALL SELECT  7, 'a', 2, 1 FROM DUAL
UNION ALL SELECT  8, 'a', 2, 1 FROM DUAL
UNION ALL SELECT  9, 'b', 1, 1 FROM DUAL
UNION ALL SELECT 10, 'b', 2, 1 FROM DUAL;

CREATE TABLE tbl2 ( id, location, invoice, payments ) AS 
          SELECT  1, 'a', 1, 1 FROM DUAL
UNION ALL SELECT  2, 'a', 1, 1 FROM DUAL
UNION ALL SELECT  3, 'a', 1, 1 FROM DUAL
UNION ALL SELECT  4, 'a', 2, 1 FROM DUAL
UNION ALL SELECT  5, 'a', 2, 1 FROM DUAL
UNION ALL SELECT  6, 'b', 1, 1 FROM DUAL
UNION ALL SELECT  7, 'b', 1, 1 FROM DUAL
UNION ALL SELECT  8, 'b', 1, 1 FROM DUAL
UNION ALL SELECT  9, 'b', 1, 1 FROM DUAL
UNION ALL SELECT 10, 'b', 1, 1 FROM DUAL;

查询 1：

这个使用相关的子查询来计算第二个表的总数：

SELECT location,
       invoice,
       SUM( payments ) AS total_payments_1,
       COALESCE( (SELECT SUM( payments )
                  FROM   tbl2 i
                  WHERE  o.location = i.location
                     AND o.invoice  = i.invoice),
                 0 ) AS total_payments_2
FROM   tbl1 o
GROUP BY
       location,
       invoice
ORDER BY
       location,
       invoice

结果：

| LOCATION | INVOICE | TOTAL_PAYMENTS_1 | TOTAL_PAYMENTS_2 |
|----------|---------|------------------|------------------|
|        a |       1 |                5 |                3 |
|        a |       2 |                3 |                2 |
|        b |       1 |                1 |                5 |
|        b |       2 |                1 |                0 |

查询 2：

这个使用命名子查询来预先计算表 1 的总数，然后LEFT OUTER JOIN对第二个表执行 a 并将表 1 的总数包含在组中。

如果没有任何索引，从解释计划来看，查询 1 似乎效率更高，但您的索引可能意味着优化器找到了更好的计划。

WITH tbl1_sums AS (
  SELECT location,
         invoice,
         SUM( payments ) AS total_payments_1
  FROM   tbl1
  GROUP BY
         location,
         invoice
)
SELECT t1.location,
       t1.invoice,
       t1.total_payments_1,
       COALESCE( SUM( t2.payments ), 0 ) AS total_payments_2
FROM   tbl1_sums t1
       LEFT OUTER JOIN 
       tbl2 t2
       ON (    t1.location = t2.location
           AND t1.invoice = t2.invoice)
GROUP BY
       t1.location,
       t1.invoice,
       t1.total_payments_1
ORDER BY
       t1.location,
       t1.invoice

结果：

| LOCATION | INVOICE | TOTAL_PAYMENTS_1 | TOTAL_PAYMENTS_2 |
|----------|---------|------------------|------------------|
|        a |       1 |                5 |                3 |
|        a |       2 |                3 |                2 |
|        b |       1 |                1 |                5 |
|        b |       2 |                1 |                0 |
于 2013-12-12T20:31:08.310 回答