SQL-输出得分范围括号内的平均满意度/晋升次数

到目前为止，我已经开发了一个查询，以输出分数范围内每个员工的计数（从评估表中得出的评估分数）。我正在努力弄清楚如何对输出的这些员工范围执行数学函数。有必要从我的员工表中提取信息，以统计最近5年/不到5年的晋升人数以及满意度水平。在有关范围的查询下方找到附件

SELECT t.range AS [0-1], count(Employee_ID) AS [Number of Employees]
FROM (
SELECT CASE 
WHEN score BETWEEN 0 AND 0.25 THEN ‘0-0.25’
WHEN score BETWEEN 0.25 AND 0.5 THEN ‘0.25-0.5’
WHEN score BETWEEN 0.5 AND 0.75 THEN ‘0.5-0.75’
WHEN score BETWEEN 0.75 AND 1 THEN ‘0.75-1’
FROM Evaluation)
GROUP BY t.range;

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这就是你想做的吗

SELECT CASE 
WHEN score BETWEEN 0 AND 0.25 THEN ‘0-0.25’
WHEN score BETWEEN 0.25 AND 0.5 THEN ‘0.25-0.5’
WHEN score BETWEEN 0.5 AND 0.75 THEN ‘0.5-0.75’
WHEN score BETWEEN 0.75 AND 1 THEN ‘0.75-1’ end AS range,
count(Employee_ID) AS [Number of Employees]
FROM Evaluation 
GROUP BY CASE 
WHEN score BETWEEN 0 AND 0.25 THEN ‘0-0.25’
WHEN score BETWEEN 0.25 AND 0.5 THEN ‘0.25-0.5’
WHEN score BETWEEN 0.5 AND 0.75 THEN ‘0.5-0.75’
WHEN score BETWEEN 0.75 AND 1 THEN ‘0.75-1’ end 

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您正在使用子查询，所以请继续使用它。您只需要正确的别名：

SELECT e.range AS [0-1], count(*) AS [Number of Employees]
FROM (SELECT (CASE WHEN score BETWEEN 0 AND 0.25 THEN '0-0.25'
                   WHEN score BETWEEN 0.25 AND 0.5 THEN '0.25-0.5'
                   WHEN score BETWEEN 0.5 AND 0.75 THEN '0.5-0.75'
                   WHEN score BETWEEN 0.75 AND 1 THEN '0.75-1'
              END) as range
      FROM Evaluation
     ) e
GROUP BY e.range;

注:

• 子查询需要一个别名。我用“e”表示“评估”
• 卷曲单引号必须是直单引号（这可能是剪切粘贴问题）

• EmployeeId

  未在外部查询中定义。我认为

  COUNT（*）

  可能是等效的

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您的派生表（子查询）返回1列（未命名），但在顶层突然有2列。您的案例语句缺少一个“结束”标记，您正在使用的数据库中的问题。这正是我要找的！