SELECT Task_Entry_Icode, Task_Master_Icode, Work_Progress,SUM(A.Logged_Hours) as logged,B.*
FROM task_entry一个INNER JOIN task_master B on A.Task_Master_Icode = B.Task_Icode
WHERE Task_Entry_Icode IN(
SELECT MAX(Task_Entry_Icode)
FROM task_entry
GROUP BY Task_Master_Icode
);
答案 0 :(得分:1)
SELECT * FROM task_master INNER JOIN task_entry B ON A.Task_Icode = B.Task_Master_Icode 在哪里Created_On IN( SELECT MAX(Created_On) FROM task_entry GROUP BY Task_Master_Icode )和A. A.Task_Created_By =' 7';
这适合我,但如果我使用sum()它只返回一个记录
答案 1 :(得分:0)
SELECT
B.Task_Entry_Icode,
B.Task_Master_Icode,
(SELECT Work_Progress FROM task_entry WHERE Task_Master_Icode=B.Task_Master_Icode ORDER BY B.Created_On DESC LIMIT 1 OFFSET 0),
SUM(B.Logged_Hours)
FROM
task_master A INNER JOIN task_entry B ON A.Task_Icode=B.Task_Master_Icode
GROUP BY
B.task_master_icode;