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我想得到每个用户和他们最后一次付款。我在这里有 2 张桌子users和finances。我尝试添加groupby并得到了我想要的结果，但是它获得了另一个表中最旧的记录。有谁知道我怎么能做到这一点？

我的第一个查询

SELECT users.name, users.email, users.phone, users.parent_id, users.section_id, finances.amount, finances.description, schoolyears.name, finances.date 
from users 
JOIN finances on users.id = finances.user_id 
JOIN schoolyears on users.school_id = schoolyears.school_id 
ORDER BY finances.date DESC;

结果我得到

+-----------------+------------------------------+--------------------+-----------+------------+--------+--------------+--------------+------------+
| name            | email                        | phone              | parent_id | section_id | amount | description  | name         | date       |
+-----------------+------------------------------+--------------------+-----------+------------+--------+--------------+--------------+------------+
| Madelynn Stokes | moore.dominic@cartwright.com | +63 (971) 659-8143 |        10 |       NULL | 1000   | New Payables | SY-2019-2020 | 2019-11-14 |
| Annamarie Morar | emile99@hotmail.com          | (0997) 212-7919    |         3 |       NULL | 500    | New Pays     | SY-2019-2020 | 2019-11-14 |
| Madelynn Stokes | moore.dominic@cartwright.com | +63 (971) 659-8143 |        10 |       NULL | 5000   | Old Payables | SY-2019-2020 | 2019-11-13 |
| Annamarie Morar | emile99@hotmail.com          | (0997) 212-7919    |         3 |       NULL | 200    | Old Pays     | SY-2019-2020 | 2019-11-13 |
+-----------------+------------------------------+--------------------+-----------+------------+--------+--------------+--------------+------------+

我只想要其他表中的最新记录。新应付款和新支付。

我尝试了第二个查询

SELECT users.name, users.email, users.phone, users.parent_id, users.section_id, finances.amount, finances.description, schoolyears.name, finances.date 
from users 
JOIN finances on users.id = finances.user_id 
JOIN schoolyears on users.school_id = schoolyears.school_id 
GROUP BY users.id 
ORDER BY finances.date DESC;

它有效，但我得到了最古老的记录。

+-----------------+------------------------------+--------------------+-----------+------------+--------+--------------+--------------+------------+
| name            | email                        | phone              | parent_id | section_id | amount | description  | name         | date       |
+-----------------+------------------------------+--------------------+-----------+------------+--------+--------------+--------------+------------+
| Madelynn Stokes | moore.dominic@cartwright.com | +63 (971) 659-8143 |        10 |       NULL | 5000   | Old Payables | SY-2019-2020 | 2019-11-13 |
| Annamarie Morar | emile99@hotmail.com          | (0997) 212-7919    |         3 |       NULL | 200    | Old Pays     | SY-2019-2020 | 2019-11-13 |
+-----------------+------------------------------+--------------------+-----------+------------+--------+--------------+--------------+------------+
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1 回答 1

2

你可以试试下面 -

演示

SELECT users.name, users.email, users.phone, users.parent_id, users.section_id, finances.amount, finances.description, schoolyears.name, finances.dateval 
from users 
JOIN finances on users.id = finances.user_id 
JOIN schoolyears on users.school_id = schoolyears.school_id 
where finances.dateval=
(select max(dateval) from finances f where finances.user_id=f.user_id)
于 2019-11-14T08:03:40.287 回答