Question

很难描述我想要的东西，但我会尝试：首先，我有2个表：员工和工资。我通过这个SQL命令加入了2个表：

"SELECT employee.id, employee.employee_name, employee.image, salary.year, salary.month, salary.day, salary.total_salary, salary.id, salary.created
                        FROM employee
                 JOIN salary ON salary.employee_id=employee.id"

获取我插入的所有值，执行后，我得到：

Created              name    year     month   day(working)       total_salary
------------------------------------------------------------------------------
2016-12-17 20:41      A      2016       1         31              1550000
2016-12-17 21:49      A      2016       1         20              2000000
2016-12-17 22:49      A      2016       2         20              2000000
2016-12-18 22:43      B      2016       2         15              1550000
2016-12-18 23:43      B      2016       2         10              1000000
2016-12-18 23:49      B      2016       2         1                100000

之后，我希望每月获得最新值，所以我将此命令添加到上面的sql结尾：

 GROUP BY employee.employee_name, salary.year, salary.month"

得到了这个：

Created              name    year     month   day(working)       total_salary
------------------------------------------------------------------------------
2016-12-17 20:41      A      2016       1         31              1550000
2016-12-17 22:49      A      2016       2         20              2000000
2016-12-18 22:43      B      2016       2         15              1550000

什么查询会返回以下结果？

Created              name    year     month   day(working)       total_salary
------------------------------------------------------------------------------
2016-12-17 21:49      A      2016       1         20              2000000
2016-12-17 22:49      A      2016       2         20              2000000
2016-12-18 23:49      B      2016       2         1                100000

Answer 1

在分组之前，您应该按照最新条目的创建日期对条目进行排序：

ORDER BY salary.created DESC

更改第一个查询（您必须删除或替换其中一个ID，因为您将有两个具有相同名称的列）：

SELECT employee.id as employee_id, employee.employee_name, employee.image, salary.year, salary.month, salary.day, salary.total_salary, salary.id, salary.created 
FROM employee JOIN salary ON salary.employee_id=employee.id 
ORDER BY salary.created DESC

首先显示最后创建的内容。然后像你一样对结果进行分组：

SELECT * FROM (
    SELECT employee.id as employee_id, employee.employee_name, employee.image, salary.year, salary.month, salary.day, salary.total_salary, salary.id, salary.created
    FROM employee
    JOIN salary ON salary.employee_id=employee.id 
    ORDER BY salary.created DESC 
) as employee_salary 
GROUP BY employee_salary.employee_name, employee_salary.year, employee_salary.month

Answer 2

我的出发点是在您加入之后，我刚刚创建了一个将数据作为您的第一个结果的表格。

使用此查询：

SELECT created, name, year, month, day, total_salary FROM 
( SELECT * FROM employees
  ORDER BY created DESC
) AS sub
GROUP BY name, month, year;

我得到了你要求的结果：

Created              name    year     month   day(working)       total_salary
------------------------------------------------------------------------------
2016-12-17 21:49      A      2016       1         20              2000000
2016-12-17 22:49      A      2016       2         20              2000000
2016-12-18 23:49      B      2016       2         1                100000

基本上诀窍是处理订购，然后再将它们打包成 GROUP BY 的组所以我做了一个只处理排序的子查询。这post帮助我弄清楚如何做到这一点。

这是一个Fiddle，您可以在其中进行游戏并根据您的起始查询进行调整。

如何过滤Mysql的结果

2 个答案: