Mysql 同一表上的多个左联接

我有两张桌子。 第一个表=>成员{member_id，name，active} 第二个表=>储蓄{储蓄id，会员id，月，年，金额，类型，支付}

成员表

+-----------+--------+--------+
| member_id | name   | active |
+-----------+--------+--------+
|       105 | Andri  | 1      |
|       106 | Steve  | 1      |
|       110 | Soraya | 1      |
|       111 | Eva    | 1      |
|       112 | Sonia  | 1      |
+-----------+--------+--------+

储蓄表

+------------+-----------+-------+------+--------+------+------+
| savings_id | member_id | month | year | amount | type | paid |
+------------+-----------+-------+------+--------+------+------+
|          1 |       120 |  NULL | NULL | 150000 |    1 | 1    |
|         14 |       105 |     7 | 2014 |  80000 |    2 | 1    |
|         15 |       105 |     7 | 2014 |  25000 |    3 | 1    |
|         16 |       105 |     7 | 2014 |  60000 |    4 | 1    |
|         17 |       105 |     7 | 2014 | 100000 |    5 | 1    |
|         18 |       106 |     7 | 2014 |  80000 |    2 | 1    |
|         19 |       106 |     7 | 2014 |  25000 |    3 | 1    |
|         20 |       106 |     7 | 2014 |  60000 |    4 | 1    |
|         21 |       106 |     7 | 2014 | 100000 |    5 | 1    |
|         31 |       110 |     7 | 2014 |  25000 |    3 | 1    |
|         32 |       110 |     7 | 2014 |  60000 |    4 | 1    |
|         33 |       110 |     7 | 2014 | 100000 |    5 | 1    |
|         34 |       111 |     7 | 2014 |  80000 |    2 | 1    |
|         35 |       111 |     7 | 2014 |  25000 |    3 | 1    |
|         36 |       111 |     7 | 2014 |  60000 |    4 | 1    |
|         37 |       111 |     7 | 2014 | 100000 |    5 | 1    |
|         38 |       112 |     7 | 2014 |  80000 |    2 | 1    |
|         39 |       112 |     7 | 2014 |  25000 |    3 | 1    |
|         40 |       112 |     7 | 2014 |  60000 |    4 | 1    |
|         41 |       112 |     7 | 2014 | 100000 |    5 | 1    |
|         85 |       105 |     7 | 2015 |  80000 |    2 | 1    |
|         86 |       105 |     7 | 2015 |  25000 |    3 | 1    |
|         87 |       105 |     7 | 2015 |  60000 |    4 | 1    |
|         88 |       105 |     7 | 2015 | 100000 |    5 | 1    |
|         89 |       106 |     7 | 2015 |  80000 |    2 |      |
|         90 |       106 |     7 | 2015 |  25000 |    3 |      |
|         91 |       106 |     7 | 2015 |  60000 |    4 |      |
|         92 |       106 |     7 | 2015 | 100000 |    5 |      |
|        101 |       110 |     7 | 2015 |  80000 |    2 |      |
|        102 |       110 |     7 | 2015 |  25000 |    3 |      |
|        103 |       110 |     7 | 2015 |  60000 |    4 |      |
|        104 |       110 |     7 | 2015 | 100000 |    5 |      |
|        105 |       111 |     7 | 2015 |  80000 |    2 | 1    |
|        106 |       111 |     7 | 2015 |  25000 |    3 | 1    |
|        107 |       111 |     7 | 2015 |  60000 |    4 | 1    |
|        108 |       111 |     7 | 2015 | 100000 |    5 | 1    |
|        109 |       112 |     7 | 2015 |  80000 |    2 |      |
|        110 |       112 |     7 | 2015 |  25000 |    3 |      |
|        111 |       112 |     7 | 2015 |  60000 |    4 |      |
|        144 |       110 |     7 | 2014 |  50000 |    1 | 1    |
+------------+-----------+-------+------+--------+------+------+

当会员进行储蓄时，他们可以选择5种类型的储蓄，我想做的是列出会员及其所有储蓄

这是mysql查询

SELECT m.member_id, name, 
SUM(s1.amount) as savings1,
SUM(s2.amount) as savings2,
SUM(s3.amount) as savings3,
SUM(s4.amount) as savings4,
SUM(s5.amount) as savings5
FROM members m
LEFT JOIN savings s1 ON s1.member_id = m.member_id AND s1.type = 1 AND s1.paid = 1
LEFT JOIN savings s2 ON s2.member_id = m.member_id AND s2.type = 2 AND s2.paid = 1
LEFT JOIN savings s3 ON s3.member_id = m.member_id AND s3.type = 3 AND s3.paid = 1
LEFT JOIN savings s4 ON s4.member_id = m.member_id AND s4.type = 4 AND s4.paid = 1
LEFT JOIN savings s5 ON s5.member_id = m.member_id AND s5.type = 5 AND s5.paid = 1
WHERE 
active = 1
GROUP BY m.member_id

这是输出

+-----------+--------+----------+----------+----------+----------+----------+
| member_id | name   | savings1 | savings2 | savings3 | savings4 | savings5 |
+-----------+--------+----------+----------+----------+----------+----------+
|       105 | Andri  |     NULL |  1280000 |   400000 |   960000 |  1600000 |
|       106 | Steve  |     NULL |    80000 |    25000 |    60000 |   100000 |
|       110 | Soraya |    50000 |     NULL |    25000 |    60000 |   100000 |
|       111 | Eva    |     NULL |  1280000 |   400000 |   960000 |  1600000 |
|       112 | Sonia  |     NULL |    80000 |    25000 |    60000 |   100000 |
+-----------+--------+----------+----------+----------+----------+----------+

如您所见，计算不正确，例如，对于成员105，savings2应该是160K。任何关于本案例的疑问的建议

---

您可能不需要多个左联接，可以按如下方式完成

SELECT 
m.member_id,
m.name,
SUM(case when s.type= 1 then s.amount end) as savings1,
SUM(case when s.type= 2 then s.amount end) as savings2,
SUM(case when s.type= 3 then s.amount end) as savings3,
SUM(case when s.type= 4 then s.amount end) as savings4,
SUM(case when s.type= 5 then s.amount end) as savings5
FROM savings s
join members m on m.member_id = s.member_id
WHERE 
m.active = 1
GROUP BY m.member_id

这应该行得通

SELECT m.member_id, name, 

sum((case when s1.type=1 then s1.amount end)) as savings1,

sum((case when s1.type=2 then s1.amount end)) as savings2,

sum((case when s1.type=3 then s1.amount end)) as savings3,

sum((case when s1.type=4 then s1.amount end)) as savings4,

sum((case when s1.type=5 then s1.amount end)) as savings5

FROM members m

LEFT JOIN savings s1 ON s1.member_id = m.member_id 

WHERE active = 1 and s1.paid=1

GROUP BY m.member_id

---

问题是在求和之前要形成完全联接乘积。因此，如果在多个

savings

表中有多个行，则最终会出现重复。如果你愿意，你可以清楚地看到发生了什么。有两种方法可以解决这个问题

在派生表中进行所有求和：

SELECT m.member_id, name, 
s1.amount as savings1,
s2.amount as savings2,
...
FROM members m
LEFT JOIN (
    select SUM(amount) as amount, member_id
    from savings
    where type = 1 and paid = 1
    group by member_id
) s1 ON s1.member_id = m.member_id
LEFT JOIN (
    select SUM(amount) as amount, member_id
    from savings
    where type = 2 and paid = 1
    group by member_id
) s2 ON s2.member_id = m.member_id
...
WHERE active = 1
GROUP BY m.member_id

连接一次并使用条件和：

SELECT m.member_id, name, 
    SUM(CASE WHEN s.type = 1 then s.amount ELSE NULL END) as savings1,
    SUM(CASE WHEN s.type = 2 then s.amount ELSE NULL END) as savings2,
    ...
LEFT JOIN savings s s2 ON s.member_id = m.member_id AND s.paid = 1
WHERE active = 1
GROUP BY m.member_id

---

join似乎有问题。。以上答案都是不错的选择

你也可以使用透视查询

SELECT m.member_id,s1.type,
SUM(s1.amount) as savings
FROM members m
LEFT JOIN savings s1 ON s1.member_id = m.member_id  AND s1.paid = 1
WHERE active = 1
GROUP BY m.member_id ,s1.type

//在这里旋转

---

然后试着做一个旋转的家伙。。这也太好了……

按m.member\u id进行分组不是更简单吗，s.type“然后围绕成员id旋转？我认为成员id=106用于储蓄2有一些错误，实际上它变成了

80000

，您的查询将返回

160000

，您必须在

s.paid=1

这样的条件下再添加一个，以获得正确的结果。@Sadikhasan，是的，添加条件将只有

80000

，但看起来OP正在为成员105寻找

savings2。当您添加条件
s.paid=1
时，它应该是160K

，然后还可以满足用户需求，并在

CASE

函数中校正

和的输出功率sql。@Charlesliam是的，我应该提到它只起作用，因为它将连接减少到单个表。如果必须对多个表中的记录求和，则不能这样做，因为仍然会得到行的笛卡尔积。