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mysql – 如何返回SQL数学运算的结果？

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所以我最近在测试中遇到了一些更高级别的SQL问题.我只有我认为SQL中的“中级”经验,而且我已经在这一天工作了一天左右.我只是想不出来.

这是问题所在：

你有一个包含4列的表：

EmployeeID     int unique
EmployeeType   int  
EmployeeSalary int  
Created        date

目标：我需要检索具有多个条目的任何EmployeeType的最新两个EmployeeSalary之间的差异.它必须在一个语句中完成(嵌套查询很好).

示例数据集：http://sqlfiddle.com/#!9/0dfc7

EmployeeID | EmployeeType | EmployeeSalary | Created
-----------|--------------|----------------|--------------------
1          | 53           | 50             | 2015-11-15 00:00:00
2          | 66           | 20             | 2014-11-11 04:20:23
3          | 66           | 30             | 2015-11-03 08:26:21
4          | 66           | 10             | 2013-11-02 11:32:47
5          | 78           | 70             | 2009-11-08 04:47:47
6          | 78           | 45             | 2006-11-01 04:42:55

所以对于这个数据集,正确的回报是：

EmployeeType | EmployeeSalary 
-------------|---------------
66           | 10
78           | 25

10来自减去EmployeeType为66的最新两个EmployeeSalary值(30 – 20).25来自减去EmployeeType为78的最新两个EmployeeSalary值(70-45).我们完全跳过EmployeeID 53,因为它只有一个价值.

这个一直在摧毁我的大脑.有线索吗？

谢谢！

解决方法:

如何制作非常简单的查询复杂？

一个有趣的方式(不是最佳表现)是：

SELECT final.EmployeeType, SUM(salary) AS difference
FROM (
  SELECT b.EmployeeType, b.EmployeeSalary AS salary
  FROM tab b
  JOIN (SELECT EmployeeType, GROUP_CONCAT(EmployeeSalary ORDER BY Created DESC) AS c
        FROM tab
        GROUP BY EmployeeType
        HAVING COUNT(*) > 1) AS sub
    ON b.EmployeeType = sub.EmployeeType
    AND FIND_IN_SET(b.EmployeeSalary, sub.c) = 1
  UNION ALL
  SELECT b.EmployeeType, -b.EmployeeSalary AS salary
  FROM tab b
  JOIN (SELECT EmployeeType, GROUP_CONCAT(EmployeeSalary ORDER BY Created DESC) AS c
        FROM tab
        GROUP BY EmployeeType
        HAVING COUNT(*) > 1) AS sub
    ON b.EmployeeType = sub.EmployeeType
    AND FIND_IN_SET(b.EmployeeSalary, sub.c) = 2
) AS final
GROUP BY final.EmployeeType;

SqlFiddleDemo

编辑：

关键点是MySQL不支持窗口函数,因此您需要使用等效代码：

例如,SQL Server中的解决方案：

SELECT EmployeeType, SUM(CASE rn WHEN 1 THEN EmployeeSalary 
                                 ELSE -EmployeeSalary END) AS difference
FROM (SELECT *,
       ROW_NUMBER() OVER(PARTITION BY EmployeeType ORDER BY Created DESC) AS rn
      FROM #tab
     ) AS sub
WHERE rn IN (1,2)
GROUP BY EmployeeType
HAVING COUNT(EmployeeType) > 1

LiveDemo

和MySQL相当：

SELECT EmployeeType, SUM(CASE rn WHEN 1 THEN EmployeeSalary 
                          ELSE -EmployeeSalary END) AS difference
FROM (
       SELECT t1.EmployeeType, t1.EmployeeSalary,
        count(t2.Created) + 1 as rn
      FROM #tab t1
      LEFT JOIN #tab t2
        ON t1.EmployeeType = t2.EmployeeType
       AND t1.Created < t2.Created
      GROUP BY t1.EmployeeType, t1.EmployeeSalary
     ) AS sub
WHERE rn IN (1,2)
GROUP BY EmployeeType
HAVING COUNT(EmployeeType) > 1;

LiveDemo2

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