Question

在oracle 11g上自学SQL，并使用员工信息数据库。

我正在尝试返回每个“办公室”员工的工作小时数和天数

但是，“工作时间”会返回每个员工轮班的正确时间，但是作为单独的行，而不是一行包含总天数和总小时数。（这意味着每行的天数= 1）

SELECT L.OFFICE_NAME as "Shop", 
(us.LAST_NAME || ' , ' || us.FIRST_NAME) AS "Employee Name",
COUNT(distinct shift.shift_date) as "WORK DAYS",
round((shift2.MaxSignOffTime-shift.MinSignOnTime)*24, 1) AS "WORK HOURS",

FROM 
LOCAL_OFFICE L
JOIN 
    Orders ord
    ON L.LOCAL_OFFICE_ID = ord.LOCAL_OFFICE_ID
JOIN  
    USERS us
    ON
     us.USER_ID = ord.ASSIGNED_TO_USER_ID
join
    (SELECT min(act_sign_dt) as MinSignOnTime, USER_ID, shift_token 
FROM CLIENT_SIGN 
WHERE
    BEG_OF_SHIFT = 'Y'
GROUP BY shift_date, user_id, shift_token)  shift on shift. user_id = ord.assigned_to_user_id
join
    (SELECT max(act_sign_dt) as MaxSignOffTime, USER_ID, shift_token 
FROM CLIENT_SIGN 
WHERE
    end_OF_SHIFT = 'Y'
GROUP BY shift_date, user_id, shift_token)  shift2 on shift2.shift_token = shift.shift_token

GROUP BY OFFICE_NAME, LAST_NAME, FIRST_NAME,shift.shift_date, (shift2.MaxSignOffTime-shift.MinSignOnTime)
ORDER BY OFFICE_NAME, "Employee Name"

我愿意回答任何需要的问题，但我根本无法理解为什么不为每个办公地点的每位员工分成一行

Answer 1

试试这个......

在SELECT：

中放置一个Sum（）

round((shift2.MaxSignOffTime-shift.MinSignOnTime)*24, 1)

并删除此内容：

(shift2.MaxSignOffTime-shift.MinSignOnTime)

来自GROUP BY子句的

自学sql，分组结果错误

1 个答案: