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我对子查询有疑问。所以我想列出总工作时间大于所有员工平均工作时间的员工。但是使用此代码，它会将一名员工的工作时间与另一名员工每小时的工作时间进行比较。所以这是总结：员工总工作时间>所有员工总工作时间的平均值-> WHERE hrs_worked > AVG(SUM(HRS_WORKED)) 这就是我想要的。

SELECT employee.emp_id,
       employee.emp_name,
       SUM(action.hrs_worked) AS "Total Hours Worked"
FROM action
INNER JOIN allocation ON action.act_id = allocation.act_id
AND action.emp_id = allocation.emp_id
INNER JOIN employee ON allocation.emp_id = employee.emp_id
WHERE action.hrs_worked >
    (SELECT AVG(action.hrs_worked)
     FROM action)
GROUP BY employee.emp_id,
         employee.emp_name
ORDER BY SUM(action.hrs_worked) DESC
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2 回答 2

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应该是这样的：

SELECT * 
FROM  Orders
........... 
WHERE Price > (
           SELECT AVG(sum_pr) 
           FROM (
              SELECT SUM(Price) as sum_pr 
              FROM Orders 
              GROUP BY Price
                 )x
               )
于 2015-04-20T10:46:41.440 回答
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让我们把员工姓名排除在外。总工时为：

SELECT a.emp_id, SUM(a.hrs_worked) AS TotalHoursWorked
FROM action a 
GROUP by a.emp_id;

要获得大于平均值的值：

SELECT a.emp_id, SUM(a.hrs_worked) AS TotalHoursWorked
FROM action a 
GROUP by a.emp_id;
HAVING TotalHoursWorked > (SELECT avg(TotalHoursWorked)
                           FROM (SELECT a.emp_id, SUM(a.hrs_worked) AS TotalHoursWorked
                                 FROM action a 
                                 GROUP by a.emp_id
                                ) a2
                           );

您可以使用它join来引入您想要的其他字段。

根据您的查询，我看不出allocation是如何使用的。

于 2015-04-20T10:54:54.513 回答