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Question

我这几天一直在为这个查询苦苦挣扎。我正在使用 PHP/MySQL。

这是一个系统，客户可以通过我正在为其开发的公司发送货运以进行收集或交付。我正在尝试编写的查询将收集为客户进行的收集和交付，并向他们发送他们当天花费的摘要。

下面有两个结果集，我需要将它们“连接在一起”才能发送报告。

（不要担心细节，问题在于连接 :)）

交付结果集

此结果是公司代表所列客户交付的交付总额列表（按天分组）。

summary_day, num_cons, num_spaces, customer_code 
2011-07-20      1       3           COB001P
2011-07-20      1       3           FAI001P
2011-07-20      2       2           FRE001P
2011-07-20      2       2           MIN001P
2011-07-20      17      29          NOR001P
2011-07-20      50      79          PAL001P
2011-07-20      1       1           PAR001P
2011-07-20      1       1           POT002P
2011-07-20      6       7           RHY001P
2011-07-20      9       13          TDG001P
2011-07-20      18      23          UPN001P

集合结果集

与上述类似，但反过来，此结果是公司代表每个客户收集的总计列表。

2011-07-20  9   15  ARR001P
2011-07-20  1   1   BAC002P
2011-07-20  1   1   BLA001P
2011-07-20  4   6   CAR003P
2011-07-20  2   2   DIS001P
2011-07-20  2   2   DOV001P
2011-07-20  1   1   DRY001P
2011-07-20  1   1   ECC001P
2011-07-20  3   5   FAI001P
2011-07-20  2   2   INV001P
2011-07-20  2   2   MIN001P
2011-07-20  3   3   PAL001P
2011-07-20  1   1   QUA002P
2011-07-20  1   1   TEC002P
2011-07-20  1   1   THE006P
2011-07-20  7   7   WIL005P

问题

我正在努力使用 JOIN 来合并这两个结果集。

理想情况下，最终结果集应该非常标准，包含以下列：

summary_day, customer_code, num_deliveries, num_collections

数字字段将是每个结果集中的 num_spaces 列。如果客户同时有收货和发货记录，则显示这两个数字。如果他们有一个没有另一个，我使用 COALESCE 来期望一列为 NULL，并相应地将其设置为 0。

我尝试在 customer_code 字段上使用 RIGHT OUTER JOIN，希望这会产生预期的结果，但我得到的唯一结果是：

2011-07-20, ARR001P, 0, 15
2011-07-20, BAC002P, 0, 1
2011-07-20, BLA001P, 0, 1
2011-07-20, CAR003P, 0, 6
2011-07-20, DIS001P, 0, 2
2011-07-20, DOV001P, 0, 2
2011-07-20, DRY001P, 0, 1
2011-07-20, FAI001P, 3, 5
2011-07-20, INV001P, 0, 2
2011-07-20, MIN001P, 2, 2
2011-07-20, PAL001P, 79, 3
2011-07-20, QUA002P, 0, 1
2011-07-20, TEC002P, 0, 1
2011-07-20, THE006P, 0, 1
2011-07-20, WIL005P, 0, 7

如您所见，结果集似乎只返回有收货的客户，这没关系，但我还需要查看有货但没有收货的客户。

例如，客户 NOR001P 属于交付结果集中但不在收款结果集中...

在这种情况下是否需要 FULL OUTER JOIN？如果是这样，考虑到 MySQL 不支持 FULL OUTER JOIN，我该如何解决这个问题？

感谢您抽出宝贵时间阅读。

完整解决方案

感谢 CResults 的回答，完整的解决方案如下... 可以看到交付和收集结果集实际上是由子查询组成的，所以这有点头疼！

set @summary_day = '2011-07-20';
select summary_day, customer_code, sum(num_deliveries) as pallet_deliveries, sum(num_collections) as pallet_collections
from
(
        select d.summary_day, d.customer_code, d.num_spaces as num_deliveries, 0 as num_collections from
            (select 
                        @summary_day as summary_day, /* change variable */
                        count(*) as num_cons,
                        sum( coalesce(micro_qty,0) + coalesce(quarter_qty,0) + coalesce(half_qty,0) + coalesce(full_qty,0) + coalesce(ceil(vlu_qty),0) ) as num_spaces,
                        pc.customer_code 
                    from pallet_routes pr
                    inner join pallet_consignments pc on pc.route_id = pr.route_id
                    where pr.date = @summary_day /* today */
                    and pc.type = 'D'
                    group by customer_code
            ) as d
        union all
        select c.summary_day, c.customer_code, 0 as num_deliveries, num_spaces as num_collections from
            (select 
                    @summary_day as summary_day, /* change variable */
                    count(*) as num_cons,
                    sum(coalesce(micro_qty,0) + coalesce(quarter_qty,0) + coalesce(half_qty,0) + coalesce(full_qty,0) + coalesce(ceil(vlu_qty),0)) as num_spaces,
                    pc.customer_code 
                from pallet_routes pr
                inner join pallet_consignments pc on pc.route_id = pr.route_id
                where pr.date = DATE_SUB(@summary_day, INTERVAL 1 DAY) /* Yesterday */
                and pc.type = 'C'
                group by customer_code
            ) as c
) as pallet_summaries
group by summary_day, customer_code

Answer 1

SELECT D.summary_day
     , D.customer_code
     , D.num_deliveries
     , COALESCE(C.num_collections, 0) AS num_collections
FROM Deliveries AS D
  LEFT JOIN Collections AS C
    ON D.customer_code = C.customer_code
    AND D.summary_day = C.summary_day

UNION ALL

SELECT C.summary_day
     , C.customer_code
     , 0
     , C.num_collections
FROM Deliveries AS D
  RIGHT JOIN Collections AS C
    ON D.customer_code = C.customer_code
    AND D.summary_day = C.summary_day
WHERE D.summary_day IS NULL

ORDER BY summary_day
       , customer_code

Answer 2

完全未经测试，但试一试

Select 
    Date, CustCode, Sum(Num_Cons) as Num_Cons, Sum(Num_Cols) as Num_Cols
from 
(    Select Date, CustCode, Num_Cons, 0 as Num_Cols From Consignments
    UNION ALL
    Select Date, CustCode, 0 as Num_Cons, Num_Cols From Collections
) parcels
group by Date, CustCode