使用聚集和MAX函数

我在与下面的查询问题，在SQL Server的工作。使用聚集和MAX函数

SELECT 
     emp_id= CASE employee_id 
     WHEN '' 
      THEN RTRIM(last_name) + '_' + RTRIM(first_name) 
      + '_' + RTRIM(gender) + '_' 
      + RTRIM(race_ethnicity_code) + '_' 
      + RTRIM(high_degree_code) + '_' + RTRIM(position_code) + '_' 
      + RTRIM(assignment_code) 
      ELSE employee_id 
      END , 
     last_name, first_name, 
     assign_perc, 
      assignment_num, 
     CAST((total_salary)AS NUMERIC (18,2))* CAST((assign_perc) AS NUMERIC (18,2)) AS salary, 
     total_salary 
FROM employee 
ORDER BY last_name, first_name, district_name

我的脚本是一个简单的列提取，当emp_id为null时，通过case语句创建一个唯一的键。我遇到的问题是，当这个人有多个分配时，将assign_perc与total_Salary相乘，并且当该销售人员仅列出一次时获得最高工资。例如 - 我的预计业绩：

enter image description here

约翰·史密斯是唯一被列具有一种分配的兼职工人只有一次，所以他assign_perc将小于1但我仍然需要最高的薪水，而不是总计（assign_perc * total_salary）。感谢您的帮助。

来源

2013-04-18 Tone

您的结果与SQL不一致。对于约翰史密斯，薪水和总薪水有相同的价值，但薪水应根据您的查询0.75 * total_salary。 –

我提供的结果是我需要的。我的查询目前正在生产（.75 * 10400）= 7800 - 约翰史密斯的薪水。 – Tone

解决方案将非常麻烦，因为您没有简单的方法来确定特定员工的条目数量，而无需具体的employee_id。 – PinnyM

如上所述，这是过于凌乱，因为你需要使用CASE语句来确定虚拟employee_id。这将是多简单，如果你可以重构这个CASE语句到一个UDF，或者结果在表存储在上帝的份！

要澄清一下，这是要干什么 - 你想加入employee表上包含了每个员工计数的表。伯爵表是这样的：

SELECT employee_id, COUNT(*) AS employee_count 
FROM employee 
GROUP BY employee_id

将它们连接在一起，应该是这样的：

SELECT ... 
FROM employee 
JOIN (SELECT employee_id, COUNT(*) AS employee_count 
     FROM employee 
     GROUP BY employee_id) ec 
ON employee.employee_id = ec.employee_id

你计算现在的薪水将成为：

 CASE 
      WHEN ec.employee_count > 1 
       THEN CAST((total_salary)AS NUMERIC (18,2))* CAST((assign_perc) AS NUMERIC (18,2)) 
      ELSE total_salary 
     END AS salary,

这里是完整的查询

SELECT 
     CASE employee.employee_id 
     WHEN '' 
      THEN RTRIM(last_name) + '_' + RTRIM(first_name) 
      + '_' + RTRIM(gender) + '_' 
      + RTRIM(race_ethnicity_code) + '_' 
      + RTRIM(high_degree_code) + '_' + RTRIM(position_code) + '_' 
      + RTRIM(assignment_code) 
      ELSE employee.employee_id 
      END AS emp_id, 
     last_name, first_name, 
     assign_perc, 
     assignment_num, 
     CASE 
      WHEN ec.employee_count > 1 
       THEN CAST((total_salary)AS NUMERIC (18,2))* CAST((assign_perc) AS NUMERIC (18,2)) 
      ELSE total_salary 
     END AS salary, 
     total_salary 
FROM employee 
JOIN (SELECT CASE employee.employee_id 
     WHEN '' 
      THEN RTRIM(last_name) + '_' + RTRIM(first_name) 
      + '_' + RTRIM(gender) + '_' 
      + RTRIM(race_ethnicity_code) + '_' 
      + RTRIM(high_degree_code) + '_' + RTRIM(position_code) + '_' 
      + RTRIM(assignment_code) 
      ELSE employee.employee_id 
      END AS employee_id, COUNT(*) employee_count 
     FROM employee 
     GROUP BY CASE employee.employee_id 
     WHEN '' 
      THEN RTRIM(last_name) + '_' + RTRIM(first_name) 
      + '_' + RTRIM(gender) + '_' 
      + RTRIM(race_ethnicity_code) + '_' 
      + RTRIM(high_degree_code) + '_' + RTRIM(position_code) + '_' 
      + RTRIM(assignment_code) 
      ELSE employee.employee_id 
      END) ec 
    ON CASE employee.employee_id 
     WHEN '' 
      THEN RTRIM(last_name) + '_' + RTRIM(first_name) 
      + '_' + RTRIM(gender) + '_' 
      + RTRIM(race_ethnicity_code) + '_' 
      + RTRIM(high_degree_code) + '_' + RTRIM(position_code) + '_' 
      + RTRIM(assignment_code) 
      ELSE employee.employee_id 
      END = ec.employee_id 
ORDER BY last_name, first_name, district_name

012：在地方“雇员标识”的代滔天的CASE语句

来源

2013-04-18 15:41:27 PinnyM

使用聚集和MAX函数

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