MySQL:具有JOIN的SUM()返回不正确的值 [英] MySQL: SUM() with JOIN returns incorrect values

问题描述

我正在尝试为表中的每个用户获取一个SUM()，但是MySQL返回的值错误.

I am trying to fetch a SUM() for each user in a table, but MySQL is returning the wrong values.

这是它的外观( http://sqlfiddle.com/#!2 /7b988/4/0 )

user    amount
110     20.898319244385
114     43.144836425781
115     20.487638473511
116     26.07483291626
117     93.054000854492

但这就是它的外观( http://sqlfiddle.com/#！ 2/7b988/2/0 )

user    amount
110     167.186554
114     129.434509
115     143.413469
116     208.598663
117     744.432007

这是我要运行的查询:

SELECT 
    blocks.user_id, 
    SUM(payout_history.amount) as amount
FROM blocks
LEFT JOIN payout_history
ON blocks.user_id = payout_history.user_id
WHERE confirms > 520
GROUP BY blocks.user_id

我在做什么错了?

推荐答案

尝试以下查询:

SELECT bl.user_id, SUM( ph.amount ) PAIDOUT
FROM (
   SELECT distinct blocks.user_id 
   FROM blocks
   WHERE confirms > 520
) bl
LEFT JOIN  payout_history ph
ON bl.user_id = ph.user_id
GROUP BY ph.user_id
;

SQLFiddle-> http://sqlfiddle.com/#!2/7b988/48

SQLFiddle --> http://sqlfiddle.com/#!2/7b988/48

---编辑---说明查询的工作方式(或更确切地说，查询为什么不工作)----

--- EDIT --- an explanation how the query works (or rather why your query doesn't work) ----

查看预期结果，似乎查询应该为每个user_id计算amount列的总和，但仅针对也在blocks表中并且具有user_id >价值超过520.
在这种情况下，简单的联接(也为左外部联接)无法工作，因为blocks表可以包含同一user_id的许多记录，例如，仅返回user_id=110的行的查询将产生以下结果: /p>

Looking at expected results it seems that the query should calculate a sum of amount column for each user_id, but only for those user_id, that are also in the blocks table, and have a blocks.confirms value grather than 520.
A simple join (also left outer join) cannot work in this case, because the blocks table can contain many records for the same user_id, for example a query that returns rows for only user_id=110 gives the following results:

SELECT *
FROM blocks
WHERE confirms > 520
      AND user_id = 110;

+ ------- + ------------ + ----------- + ------------- +
| id      | user_id      | reward      | confirms      |
+ ------- + ------------ + ----------- + ------------- +
| 0       | 110          | 20.89832115 | 521           |
| 65174   | 110          | 3.80357075  | 698           |
| 65204   | 110          | 4.41933060  | 668           |
| 65218   | 110          | 4.69059801  | 654           |
| 65219   | 110          | 4.70222521  | 653           |
| 65230   | 110          | 4.82805490  | 642           |
| 65265   | 110          | 5.25058079  | 607           |
| 65316   | 110          | 6.17262650  | 556           |
+ ------- + ------------ + ----------- + ------------- +

straight联接(和LEFT/RIGHT外部联接)以这种方式工作，它从第一个联接表中获取每个记录，并将该记录与另一个联接表中的所有行配对(组合)，从而满足联接条件.

在我们的例子中，左联接产生一个下面的结果集:

The straigh join (and LEFT/RIGHT outer join) works in this way, that takes each record from the first joinded table, and pair this record (combine it) with all rows from the other joinded table thet meet the join condition.

In our case the left join produces a below resultset:

SELECT *
FROM blocks
LEFT JOIN payout_history
ON blocks.user_id = payout_history.user_id
WHERE confirms > 520
    AND blocks.user_id = 110;
+ ------- + ------- + ----------- + -------- + --- + ------- + ----------- +
| id      | user_id | reward      | confirms | id  | user_id | amount      |
+ ------- + ------- + ----------- + -------- + --- + ------- + ----------- +
| 0       | 110     | 20.89832115 | 521      | 1   | 110     | 20.898319   |
| 65174   | 110     | 3.80357075  | 698      | 1   | 110     | 20.898319   |
| 65204   | 110     | 4.41933060  | 668      | 1   | 110     | 20.898319   |
| 65218   | 110     | 4.69059801  | 654      | 1   | 110     | 20.898319   |
| 65219   | 110     | 4.70222521  | 653      | 1   | 110     | 20.898319   |
| 65230   | 110     | 4.82805490  | 642      | 1   | 110     | 20.898319   |
| 65265   | 110     | 5.25058079  | 607      | 1   | 110     | 20.898319   |
| 65316   | 110     | 6.17262650  | 556      | 1   | 110     | 20.898319   |
+ ------- + ------- + ----------- + -------- + --- + ------- + ----------- +

现在，如果我们添加SUM( amount ) .... GROUP BY user_id，则MySql将计算上述结果集中所有amount值的总和(8行* 20.898 =〜167.184)

and now if we add SUM( amount ) .... GROUP BY user_id, MySql will calucate a sum of all amount values from the above resultset ( 8 rows * 20.898 = ~ 167.184 )

SELECT blocks.user_id, sum( amount)
FROM blocks
LEFT JOIN payout_history
ON blocks.user_id = payout_history.user_id
WHERE confirms > 520
    AND blocks.user_id = 110
GROUP BY blocks.user_id;
+ ------------ + ----------------- +
| user_id      | sum( amount)      |
+ ------------ + ----------------- +
| 110          | 167.186554        |
+ ------------ + ----------------- +

正如您在这种情况下看到的那样，联接并没有给我们期望的结果-我们需要一个名为a semi join的东西-以下是半联接的不同变体，请尝试一下:

As you see in this case the join doesn't give us desired results - we need something named a semi join - below are different variants of semi joins, give them a try:

SELECT bl.user_id, SUM( ph.amount ) PAIDOUT
FROM (
   SELECT distinct blocks.user_id 
   FROM blocks
   WHERE confirms > 520
) bl
LEFT JOIN  payout_history ph
ON bl.user_id = ph.user_id
GROUP BY ph.user_id
;


SELECT ph.user_id, SUM( ph.amount ) PAIDOUT
FROM payout_history ph
WHERE ph.user_id IN (
     SELECT user_id FROM blocks
     WHERE confirms > 520
  )
GROUP BY ph.user_id
;

SELECT ph.user_id, SUM( ph.amount ) PAIDOUT
FROM payout_history ph
WHERE EXISTS (
     SELECT 1 FROM blocks bl
     WHERE bl.user_id = ph.user_id
        AND bl.confirms > 520
  )
GROUP BY ph.user_id
;

这篇关于MySQL:具有JOIN的SUM()返回不正确的值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！