带有条件聚合的 MySQL UNION ALL (Full Join)答案

【问题标题】：MySQL UNION ALL (Full Join) with conditional aggregation带有条件聚合的 MySQL UNION ALL (Full Join)
【发布时间】：2018-03-30 00:36:59
【问题描述】：

我有三张桌子：

CREATE TABLE `Agreement` (
  `AID` bigint(20) NOT NULL AUTO_INCREMENT,
  `FLAGS` bigint(20) NOT NULL DEFAULT '0',
  PRIMARY KEY (`AID`)
);

CREATE TABLE `Assessment` (
  `ASMID` bigint(20) NOT NULL AUTO_INCREMENT,
  `AID` bigint(20) NOT NULL DEFAULT '0',
  `Amount` decimal(19,4) NOT NULL DEFAULT '0.0000',
  `Description` text,
  PRIMARY KEY (`ASMID`)
);

CREATE TABLE `Payment` (
  `RID` bigint(20) NOT NULL AUTO_INCREMENT,
  `AID` bigint(20) NOT NULL DEFAULT '0',
  `ASMID` bigint(20) NOT NULL DEFAULT '0',
  `Amount` decimal(19,4) NOT NULL DEFAULT '0.0000',
  `Description` text,
  PRIMARY KEY (`RID`)
);

我正在插入一个协议、三个评估、五个付款行，如下所述：

INSERT INTO Agreement(FLAGS) VALUES(0);
INSERT INTO Assessment(AID, Amount, Description) VALUES (1, 1200, "Rent");
INSERT INTO Assessment(AID, Amount, Description) VALUES (1, 20, "Damage - car break");
INSERT INTO Assessment(AID, Amount, Description) VALUES (1, 500, "Damage - vehicle");
INSERT INTO Payment(AID, ASMID, Amount, Description) VALUES(1, 1, 500, "Rent Fee");
INSERT INTO Payment(AID, ASMID, Amount, Description) VALUES(1, 1, 600, "Rent Fee");
INSERT INTO Payment(AID, ASMID, Amount, Description) VALUES(1, 2, 20, "Damage Fee");
INSERT INTO Payment(AID, Amount, Description) VALUES(1, 600, "Deposit Fee");
INSERT INTO Payment(AID, Amount, Description) VALUES(1, 50, "Application Fee");

当我看到数据时，它应该是这样的：

mysql> SELECT * FROM Agreement;
+-----+-------+
| AID | FLAGS |
+-----+-------+
|   1 |     0 |
+-----+-------+
1 row in set (0.00 sec)

mysql> SELECT * FROM Assessment;
+-------+-----+-----------+--------------------+
| ASMID | AID | Amount    | Description        |
+-------+-----+-----------+--------------------+
|     1 |   1 | 1200.0000 | Rent               |
|     2 |   1 |   20.0000 | Damage - car break |
|     3 |   1 |  500.0000 | Damage - vehicle   |
+-------+-----+-----------+--------------------+
3 rows in set (0.00 sec)

mysql> SELECT * FROM Payment;
+-----+-----+-------+----------+-----------------+
| RID | AID | ASMID | Amount   | Description     |
+-----+-----+-------+----------+-----------------+
|   1 |   1 |     1 | 500.0000 | Rent Fee        |
|   2 |   1 |     1 | 600.0000 | Rent Fee        |
|   3 |   1 |     2 |  20.0000 | Damage Fee      |
|   4 |   1 |     0 | 600.0000 | Deposit Fee     |
|   5 |   1 |     0 |  50.0000 | Application Fee |
+-----+-----+-------+----------+-----------------+
5 rows in set (0.00 sec)

因此，任何协议都有多个评估，都需要在不久的将来支付。它可能有多个付款，这些付款可能与评估相关（即租金）或可能不相关（即申请费）。

现在，实际上，有多个协议具有多个评估和多个付款。

现在我想要包含两个表 Assessment 和 Payment 中与 Agreement GROUPED BY 第一个协议，第二个评估相关联的所有行的结果。此外，我需要将每个评估表 Payment 表中的 AMOUNT 聚合为 PaymentsApplied，以便我们可以将其与表 Assessment 中的 Amount 比较为 AmountDue。另外，如果任何付款与任何评估无关，则不要进行汇总。结果如下所示：

+-----+-------+-----------+-----------------+--------------------+-----------------+
| AID | ASMID | AmountDue | PaymentsApplied |          ASM-Descr | PMT-Description |
+-----+-------+-----------+-----------------+--------------------+-----------------+
|   1 |     1 | 1200.0000 |       1100.0000 |               Rent |        Rent Fee |
|   1 |     2 |   20.0000 |         20.0000 | Damage - car break |     Damange Fee |
|   1 |     3 |  500.0000 |            NULL |   Damage - vehicle |            NULL |
|   1 |     0 |      NULL |        600.0000 |               NULL |     Deposit Fee |
|   1 |     0 |      NULL |         50.0000 |               NULL | Application Fee |
+-----+-------+-----------+-----------------+--------------------+-----------------+
5 Rows

我尽力解释情况。实际上，在我的应用程序查询中已经加入了 10 多个表，如协议！

欢迎任何帮助！

更新 1

我从这个查询开始，

(SELECT DISTINCT
    Payment.RID, Payment.Amount as PaymentsApplied, Payment.ASMID as PMT_ASMID, null as AmountDue, null AS ASMID
FROM Payment
    LEFT JOIN Assessment ON Assessment.ASMID=Payment.ASMID)
UNION
(SELECT DISTINCT
    null, null, null, Assessment.Amount, Assessment.ASMID
FROM Assessment
    LEFT JOIN Payment ON Payment.ASMID=Assessment.ASMID)
ORDER BY ASMID, PMT_ASMID;

这给了我结果，

+------+-----------------+-----------+-----------+-------+
| RID  | PaymentsApplied | PMT_ASMID | AmountDue | ASMID |
+------+-----------------+-----------+-----------+-------+
| NULL |            NULL |      NULL | 1200.0000 |     1 |
| NULL |            NULL |      NULL |   20.0000 |     2 |
| NULL |            NULL |      NULL |  500.0000 |     3 |
|    1 |        500.0000 |         1 |      NULL |  NULL |
|    2 |        600.0000 |         1 |      NULL |  NULL |
|    3 |         20.0000 |         2 |      NULL |  NULL |
|    4 |        600.0000 |         0 |      NULL |  NULL |
|    5 |         50.0000 |         0 |      NULL |  NULL |
+------+-----------------+-----------+-----------+-------+
8 rows in set (0.01 sec)

现在，从这一点开始，IDK 如何按评估 ID (ASMID) 聚合付款行并与 Agreement 表进行连接？

更新 2

我制作了sqlfiddle link，以防有人想尝试。

我在查询中添加了条件聚合，

(SELECT DISTINCT
    null as AmountDue,
    null AS ASMID,
    null as ASM_Descr,
    Payment.Description as PMT_Descr,
    (CASE WHEN Payment.ASMID > 0 THEN SUM(Payment.Amount) ELSE Payment.Amount END) as PaymentsApplied,
    (CASE WHEN Payment.ASMID > 0 THEN GROUP_CONCAT(Payment.RID) ELSE Payment.RID END) as PaymentList,
    Payment.ASMID as PMT_ASMID
FROM Payment
    LEFT JOIN Assessment ON Assessment.ASMID=Payment.ASMID
    GROUP BY Assessment.ASMID)
UNION ALL
(SELECT DISTINCT
    Assessment.Amount,
    Assessment.ASMID,
    Assessment.Description,
    null,
    null,
    null,
    null
FROM Assessment
    LEFT JOIN Payment ON Payment.ASMID=Assessment.ASMID
    GROUP BY Assessment.ASMID)
ORDER BY ASMID, PMT_ASMID;

这给了我，

+-----------+-------+--------------------+-------------+-----------------+-------------+-----------+
| AmountDue | ASMID | ASM_Descr          | PMT_Descr   | PaymentsApplied | PaymentList | PMT_ASMID |
+-----------+-------+--------------------+-------------+-----------------+-------------+-----------+
|      NULL |  NULL | NULL               | Deposit Fee |        600.0000 | 4           |         0 |
|      NULL |  NULL | NULL               | Rent Fee    |       1100.0000 | 1,2         |         1 |
|      NULL |  NULL | NULL               | Damage Fee  |         20.0000 | 3           |         2 |
| 1200.0000 |     1 | Rent               | NULL        |            NULL | NULL        |      NULL |
|   20.0000 |     2 | Damage - car break | NULL        |            NULL | NULL        |      NULL |
|  500.0000 |     3 | Damage - vehicle   | NULL        |            NULL | NULL        |      NULL |
+-----------+-------+--------------------+-------------+-----------------+-------------+-----------+

但是，这一行仍然缺少付款行中的一行（RID：5），我没有得到预期的结果。

【问题讨论】：

看来你需要了解一下 right join 的作用。（而不是选择那些空值。）此外，完全连接是左右连接的联合，而不是全部联合。

标签： mysql aggregate outer-join

【解决方案1】：

您似乎希望每个 AID + ASMID + PMT 描述都有一个结果行。所以：

从付款中选择并汇总。
从评估中选择。
将两者完全连接起来，因为可以有无需访问的付款和无需付款的访问。

MySQL 缺少FULL OUTER JOIN。所以写两次相同的查询，一次使用LEFT OUTER JOIN，一次使用RIGHT OUTER JOIN，然后在两个结果集上使用UNION。

select
  p.aid,
  p.asmid,
  a.amount as amount_due,
  p.payments_applied,
  a.description as asm_description,
  p.description as pmt_description
from
(
  select aid, asmid, description, sum(amount) as payments_applied
  from payment
  group by aid, asmid, description
) p
left join assessment a on a.aid = p.aid and a.asmid = p.asmid
union
select
  p.aid,
  p.asmid,
  a.amount as amount_due,
  p.payments_applied,
  a.description as asm_description,
  p.description as pmt_description
from
(
  select aid, asmid, description, sum(amount) as payments_applied
  from payment
  group by aid, asmid, description
) p
right join assessment a on a.aid = p.aid and a.asmid = p.asmid
order by aid, asmid, pmt_description;

一旦 MySQL 具有 FULL OUTER JOIN 功能，您就可以将此查询减半。

【讨论】：

你是对的关于分组的顺序！我只想问一个问题，你的查询比@JanZeiseweis 的查询效率高吗？
@rsudip90：我不知道。我喜欢在加入之前进行聚合，以便更清楚地知道发生了什么，并且在以后添加聚合时减少出错的可能性。这个查询可以使用FULL OUTER JOIN。 LEFT/RIGHT/UNION 是一个简单的解决方法。选择带/不带付款的评估（外部连接）和不带评估的付款（不连接，仅from payments）然后与UNION ALL 组合而不是UNION 应该更快。
@rsudip90 在你给我们一个包含上下文和权衡的定义之前，“更高效”没有任何意义。如果您不知道如何编写查询，请不要担心编写“高效”查询。学习如何写一个简单的。
嗨@philipxy，是的，我同意你的看法！其实我不是故意要冒犯的。但是 Jan Zeiseweis 所做的在逻辑上是有道理的，我的意思是，第一眼我就明白了这个想法，而关于 Thorsten 的答案，我真的不得不对论文进行头脑风暴（也许我还需要了解更多！）。实际上，你知道吗，我很困惑我应该去哪一个。两个查询都给出相同的结果，因此最初的目的是深入了解两者的技术差异。 :) :)
顺便说一句，@ThorstenKettner，我必须在 group by 子句的子查询中添加一个条件，即 GROUP BY AID, CASE WHEN ASMID > 0 THEN ASMID ELSE RID END; 以达到预期的结果！

【解决方案2】：

我会首先收集所有评估，将它们加入付款，然后合并所有未评估的付款：

# Assessments with payments

SELECT asm.AID,
       asm.ASMID,
       min(asm.Amount) AS AmountDue,
       SUM(pam.Amount) AS PaymentsApplied,
       asm.Description AS `ASM-Descr`,
       pam.Description AS `PMT-Descr`,
       agr.FLAGS
FROM Assessment asm
LEFT JOIN Payment pam ON pam.ASMID = asm.ASMID
JOIN Agreement agr ON agr.AID = asm.AID
GROUP BY asm.AID,
         asm.ASMID
UNION # Payments without assessments

SELECT pam.AID,
       pam.ASMID,
       NULL AS AmountDue,
       SUM(pam.Amount) AS PaymentsApplied,
       NULL AS `ASM-Descr`,
       pam.Description AS `PMT-Descr`,
       agr.FLAGS
FROM Payment pam
LEFT JOIN Assessment asm ON pam.ASMID = asm.ASMID
JOIN Agreement agr ON agr.AID = pam.AID
WHERE asm.ASMID IS NULL
GROUP BY pam.AID, pam.RID;

如果您想添加更多信息，可以包装此结果，为其命名并将更多表连接到临时结果：

SELECT payment_overview.*,
       p.name
FROM
    ( # Assessments with payments
 SELECT asm.AID,
        asm.ASMID,
        min(asm.Amount) AS AmountDue,
        SUM(pam.Amount) AS PaymentsApplied,
        asm.Description AS `ASM-Descr`,
        pam.Description AS `PMT-Descr`,
        agr.FLAGS
     FROM Assessment asm
     LEFT JOIN Payment pam ON pam.ASMID = asm.ASMID
     JOIN Agreement agr ON agr.AID = asm.AID
     GROUP BY asm.AID,
              asm.ASMID
     UNION # Payments without assessments
 SELECT pam.AID,
        pam.ASMID,
        NULL AS AmountDue,
        SUM(pam.Amount) AS PaymentsApplied,
        NULL AS `ASM-Descr`,
        pam.Description AS `PMT-Descr`,
        agr.FLAGS
     FROM Payment pam
     LEFT JOIN Assessment asm ON pam.ASMID = asm.ASMID
     JOIN Agreement agr ON agr.AID = pam.AID
     WHERE asm.ASMID IS NULL
     GROUP BY pam.AID,
              pam.RID ) AS payment_overview
JOIN Payor p ON p.AID = payment_overview.AID ;

【讨论】：

好的，谢谢您的回复。惊人的！另一个问题，我如何从协议表中获取 FLAGS。现实情况是，在此之后，我对其他 10 个表进行了多次连接，并且我还想从这些表中获取一些列
我刚刚添加了标志。
如果我还有另外 10 张桌子，我应该在协议桌子之后加入它们吗？什么是 GROUP BY 1,2？其实我没听懂！
要记住的一件事。第一部分中的付款说明可能会出现意外情况。由于您正在汇总付款金额（总和），因此您合并了多笔付款。但由于描述中没有聚合，并且它们可能不同，因此您可能会得到随机结果。
对，我知道。