Question

当我尝试执行

时，此SQL查询失败

＆＃34;当SUM（CONVERT（十进制（10,2），DATEDIFF（第二，t.Start，t.Finish）/ 3600.0））＆gt; 8＆＃34;

在我的案例陈述中。

SELECT e.EmployeeNumber, e.FirstName, e.LastName, cast(t.DateWorked as date) as DateWorked,
SUM(CONVERT(decimal(10,2),DATEDIFF(second, t.Start, t.Finish ) / 3600.0)) as HoursWorked,
SUM(CASE 
-- #Check for overtime# HoursWorked > DailyHours
WHEN SUM(CONVERT(decimal(10,2),DATEDIFF(second, t.Start, t.Finish ) / 3600.0)) > 8
    THEN 
        1
ELSE
        0
END) as moneyEarned
FROM Timesheets t, Employees e
WHERE cast(t.DateWorked as date) BETWEEN @datefrom AND @dateto
AND t.IsDeleted = 0
AND t.EmployeeId = e.EmployeeId
AND t.FarmId = @farmid
GROUP BY e.EmployeeNumber, e.FirstName, e.LastName, cast(t.DateWorked as date)

有没有办法可以访问＆＃34; HoursWorked＆＃34;作为我的Case语句中的变量？

目前我正在获得＆＃34;无法对包含聚合或子查询的表达式执行聚合函数。＆＃34;错误。

任何关于我如何以不同方式处理这一点的指示都将非常感激。

Answer 1

sum()内有sum()，但不允许这样做。据推测，你打算：

SELECT e.EmployeeNumber, e.FirstName, e.LastName,
       cast(t.DateWorked as date) as DateWorked,
       SUM(CONVERT(decimal(10,2),DATEDIFF(second, t.Start, t.Finish ) / 3600.0)) as HoursWorked,
       (CASE -- #Check for overtime# HoursWorked > DailyHours
             WHEN SUM(CONVERT(decimal(10,2),DATEDIFF(second, t.Start, t.Finish ) / 3600.0)) > 8
             THEN 1 ELSE 0
       END) as moneyEarned
. . .

您还应该学会使用正确的显式join语法。

在CASE WHEN中使用SUM时，无法对包含聚合或子查询的表达式执行聚合函数

1 个答案: