MySQL 來自三個表的“內部聯接”查詢給出錯誤結果

Question

我正在做一個 HR 項目，我需要從 mysql 數據庫中調用 3 個表值，並且需要基於以下一些標准的自定義 output：但我無法獲得 Z78E6221F6393D.456681htDB3698 我是 mysql 的新手，需要您的指導來獲得 output，如下所示。 提前感謝您的明智建議...

表員工

EMP_ID - EMP_NAME

EID001 - 湯姆

EID002 - 拉里

餐桌考勤

EMP_ID - EMP_NAME - ATTENDANCE_DATE

EID001 - 湯姆 - 20200101

EID002 - 拉里 - 20200101

EID001 - 湯姆 - 20200102

EID002 - 湯姆 - 20200103

EID002 - 湯姆 - 20200104

表 Leave_Record

EMP_ID - EMP_NAME - LEAVE_DATE(FROM) - LEAVE_DATE(TO) - DAYS - LEAVE_TYPE

EID002 - 拉里 - 20200102 - 20200102 - 1 - 休閑

EID002 - 拉里 - 20200103 - 20200103 - 1 - 生病了

EID002 - 拉里 - 20200104 - 20200104 - 1 - 年度

看着 output

EMP_ID - EMP_NAME - WORKING_DAYS - CASUAL_LEAVE - SICK_LEAVE - ANNUAL_LEAVE

EID001 -湯姆 - 4 - 0 - 0 - 0

EID002 - 拉里 - 4 - 1 - 1 - 1

我的 MySql 查詢： SELECT EMP_ID, EMP_Name, SUM(Attendance. ATTENDANCE_DATE,='') as WORKING_DAYS. SUM(Leave_Record,LEAVE_TYPE='CASUAL') as CASUAL. SUM(Leave_Record,LEAVE_TYPE=SICK) as SICK. SUM(Leave_Record,LEAVE_TYPE=ANNUAL) as ANNUAL from Employee, Attendance. Leave_Record where Attendance. EMP_ID = Employee. EMP_ID AND Employee. EMP_ID = Leave_Record. EMP_ID GROUP BY Employee. EMP_ID order by Employee. EMP_ID SELECT EMP_ID, EMP_Name, SUM(Attendance. ATTENDANCE_DATE,='') as WORKING_DAYS. SUM(Leave_Record,LEAVE_TYPE='CASUAL') as CASUAL. SUM(Leave_Record,LEAVE_TYPE=SICK) as SICK. SUM(Leave_Record,LEAVE_TYPE=ANNUAL) as ANNUAL from Employee, Attendance. Leave_Record where Attendance. EMP_ID = Employee. EMP_ID AND Employee. EMP_ID = Leave_Record. EMP_ID GROUP BY Employee. EMP_ID order by Employee. EMP_ID

Answer 1

您在兩個表中都有多個匹配的行，因此外部聚合會產生錯誤的計數。 您通常會先聚合子查詢，然后加入外部查詢：

select 
    e.emp_id, 
    e.name, 
    coalesce(a.working_days, 0) working_days, 
    coalesce(cl.casual_leave, 0) casual_leave, 
    coalesce(cl.sick_leave, 0) sick_leave
from employee e
left join (
    select emp_id, count(*) working_days from attendance group by emp_id
) a on a.emp_id = e.emp_id
left join (
    select 
        emp_id, 
        sum(case when leave_type = 'casual' then days end) casual_leave,
        sum(case when leave_type = 'sick' then days end) sick_leave
    from leave_record
    group by emp_id
) cl on cl.emp_id = e.emp_id

如果leave_record(days)始終為1 ，則第二個子查詢可以簡化為：

left join (
    select 
        emp_id, 
        sum(leave_type = 'casual') casual_leave,
        sum(leave_type = 'sick') sick_leave
    from leave_record
    group by emp_id
) cl on cl.emp_id = e.emp_id