MySQL select orderby date 返回重复值答案

【问题标题】：MySQL select orderby date returns duplicate valuesMySQL select orderby date 返回重复值
【发布时间】：2020-03-10 03:01:52
【问题描述】：

我想了解每个用户以及他们最后一次付款。我在这里有两张桌子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;

我得到的结果

+-----------------+------------------------------+--------------------+-----------+------------+--------+--------------+--------------+------------+
| 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。新应付款和新付款。

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 |
+-----------------+------------------------------+--------------------+-----------+------------+--------+--------------+--------------+------------+

【问题讨论】：

见meta.stackoverflow.com/questions/333952/…
该 GROUP BY 无效，它不会在较新的 MySQL 版本上执行（除非在兼容模式下）。您通常 GROUP BY 与您 SELECT 相同的列，除了那些是设置函数的参数的列。
您使用的是哪个 MySQL 版本？
MariaDB 10.3.1,phpMyAdmin 4.9.0.1,PHP 7.3.7

标签： mysql sql mariadb-10.3

【解决方案1】：

你可以试试下面-

DEMO

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)

【讨论】：

您确实应该在发布查询之前对其进行测试。
谢谢先生！先生，您能解释一下您做了什么吗？
@Vince，有一个语法错误兄弟 :) 而不是我使用 'on' 的地方 :)
我正在尝试将其转换为 laravel 查询构建器，但这很难哈哈