Question

我正在制作5张桌子的统计数据。我用一个客户数据制作了这个例子。

贷款

id   | status
------------
1454 | payed

付款时间表

id   | loan_id | user_client_id
-----------------------------
1456 | 1454    | 3113

payment_schedule_row

id   | payment_schedule_id | payment | payment_date
---------------------------------------------------
5013 |  1456               |    32   | 2013-11-06
5014 |  1456               |    32   | 2013-12-06
5015 |  1456               |    32   | 2013-01-05
5016 |  1456               |    32   | 2013-02-04
5017 |  1456               |    32   | 2013-03-06
5018 |  1456               |    32   | 2013-04-05
5019 |  1456               |    32   | 2013-05-05
5020 |  1456               |    32   | 2013-06-04
5021 |  1456               |    32   | 2013-07-04
5022 |  1456               |    32   | 2013-08-03
5023 |  1456               |    32   | 2013-09-02
5014 |  1456               |    32   | 2013-10-02

payment_schedule_cover

id    | payment_schedule_id |    date    | sum
----------------------------------------------
2282  | 1456                | 2013-11-08 | 34
3054  | 1456                | 2013-12-07 | 40
3776  | 1456                | 2013-01-04 | 38
4871  | 1456                | 2013-02-06 | 49
5954  | 1456                | 2013-03-06 | 40
7070  | 1456                | 2013-04-25 | 49
9029  | 1456                | 2013-05-21 | 52
10377 | 1456                | 2013-06-20 | 30
10391 | 1456                | 2013-06-21 | 30
10927 | 1456                | 2013-07-07 | 60

payment_schedule_delay

id    | payment_schedule_row_id | start_date | end_date   | delay 
----------------------------------------------------------------
1135  | 5013                    | 2013-11-07 | 2013-11-08 | 0.07
1548  | 5014                    | 2013-12-07 | 2013-12-07 | 0.03
2628  | 5016                    | 2014-02-05 | 2014-02-06 | 0.01

查询是：

SELECT period, loan_sum, covers, delay
FROM
    (SELECT MAX(EXTRACT(YEAR_MONTH FROM psc.date)) AS period, 
    (SELECT SUM(psr2.payment) FROM payment_schedule_row AS psr2 WHERE psr.payment_schedule_id = psr2.payment_schedule_id) AS loan_sum,
    (SELECT SUM(psc2.sum) FROM payment_schedule_cover AS psc2 WHERE psc.payment_schedule_id = psc2.payment_schedule_id) AS covers,
    (SELECT SUM(psd2.delay) FROM payment_schedule_delay AS psd2 WHERE psr.id = psd2.payment_schedule_row_id) AS delay
    FROM loan
    INNER JOIN payment_schedule AS ps ON ps.loan_id = loan.id
    INNER JOIN payment_schedule_row AS psr ON psr.payment_schedule_id = ps.id
    INNER JOIN payment_schedule_cover AS psc ON psc.payment_schedule_id = ps.id
    WHERE loan.status = 'payed'
    GROUP BY ps.id) AS sum_by_id
GROUP BY period

查询结果：

period  | loan_sum  | covers | delay
-----------------------------------
201407  | 384       | 422    | 0.07

除了延迟，一切都是正确的。它应该是0.11（0.07 + 0.03 + 0.01）

所以我一直试图在查询中找到错误。也许有人可以告诉我我做错了什么。

Sqlfiddle链接： http://sqlfiddle.com/#!2/21585/2

Answer 1

SELECT period, loan_sum, covers, delay
FROM
    (SELECT MAX(EXTRACT(YEAR_MONTH FROM psc.date)) AS period, 
    (SELECT SUM(psr2.payment) FROM payment_schedule_row AS psr2 WHERE psr.payment_schedule_id = psr2.payment_schedule_id) AS loan_sum,
    (SELECT SUM(psc2.sum) FROM payment_schedule_cover AS psc2 WHERE psc.payment_schedule_id = psc2.payment_schedule_id) AS covers,
    (SELECT SUM(psd2.delay) FROM payment_schedule_delay AS psd2 WHERE psr.id = psd2.payment_schedule_row_id) AS delay
    FROM loan
    INNER JOIN payment_schedule AS ps ON ps.loan_id = loan.id
    INNER JOIN payment_schedule_row AS psr ON psr.payment_schedule_id = ps.id
    INNER JOIN payment_schedule_cover AS psc ON psc.payment_schedule_id = ps.id
    INNER JOIN payment_schedule_delay AS psd ON psr.id = psd.payment_schedule_row_id
    WHERE loan.status = 'payed'
    GROUP BY ps.id) AS sum_by_id
GROUP BY period

Answer 2

SELECT period, loan_sum, covers, delay
FROM
    (SELECT MAX(EXTRACT(YEAR_MONTH FROM psc.date)) AS period, 
    (SELECT SUM(psr2.payment) FROM payment_schedule_row AS psr2 WHERE psr.payment_schedule_id = psr2.payment_schedule_id) AS loan_sum,
    (SELECT SUM(psc2.sum) FROM payment_schedule_cover AS psc2 WHERE psc.payment_schedule_id = psc2.payment_schedule_id) AS covers,
    (SELECT SUM(psd2.delay) FROM payment_schedule_delay AS psd2 WHERE psr.id IN /* IN operator will allow for multiple values like psr.id IN (5013,5014,5016) */ psd2.payment_schedule_row_id) AS delay
    FROM loan
    INNER JOIN payment_schedule AS ps ON ps.loan_id = loan.id
    INNER JOIN payment_schedule_row AS psr ON psr.payment_schedule_id = ps.id
    INNER JOIN payment_schedule_cover AS psc ON psc.payment_schedule_id = ps.id
    WHERE loan.status = 'paid'
    GROUP BY ps.id) AS sum_by_id
GROUP BY period

在汇总延迟值的行中将=更改为IN。

Answer 3

我终于从MySQL论坛得到了答案。解决问题的答案是：

......有问题...

子查询中的Group By运算符在相关的子子查询和中没有看到聚合。这些总和需要移出一个级别。
外部查询的Group By to group没有聚合;它只是作为一个Order By
像a，b，c sum（d）... group by a的查询可以返回b和c的任意结果，除非a和b之间存在严格的1：1关系，在您的子查询中看起来不太可能。
相关的子查询效率低下，正如您的两阶段连接所示;
delay相关子查询未加入任何内容

所以将相关子查询逻辑移到FROM子句，加入delay查询，修改Group By子句，我们就......

select psc.period, psc.sum, psr.payments, sum(psd.delay) as delay
from loan
join payment_schedule as ps on ps.loan_id = loan.id
join(
    select payment_schedule_id, sum(payment) as payments
    from payment_schedule_row
    group by payment_schedule_id
) as psr on psr.payment_schedule_id = ps.id
join (
    select payment_schedule_id, sum(sum) as sum, max( extract(year_month from date) ) as period
    from payment_schedule_cover
    group by payment_schedule_id
) psc on ps.id = psc.payment_schedule_id
join payment_schedule_row psr2 on ps.id = psr2.payment_schedule_id
join (
    select payment_schedule_row_id, sum(delay) as delay
    from payment_schedule_delay
    group by payment_schedule_row_id
) as psd on psr2.id = psd.payment_schedule_row_id
where loan.status = 'payed'
group by psc.period, psc.sum, psr.payments;

SQL SUM（）仅在子查询中计算一行

3 个答案: