如何从一个连接语句中的左表中获取所有行？

Question

I have two tables. One is Employee_Mstr and other is EmployeeLeaveRequest_mstr .

My data:

Answer 1

Use GROUP BY .

select t1.emp_cd, t1.emp_name, sum(t2.num_dy) from 
  Employee_Mstr t1 left join 
  EmployeeLeaveRequest_mstr t2 on t1.emp_cd = t2.emp_cd group by t1.emp_cd, t2.emp_name;

Answer 2

SELECT tab1.emp_cd, tab1.emp_name, SUM(tab2.num_dy)
FROM Employee_Mstr tab1, "2nd Table Records" tab2
where tab1.emp_cd = tab2.emp_cd
group by tab1.emp_cd, tab1.emp_name;

Should work, just set the correct table name for the 2nd table.

Answer 3

Make pk emp_cd the primary key, and then join the tales using left join . You can then use sumof to add values, use groupby before adding the values.

If you want, just check w3school.com.

Answer 4

select mstr.Emp_cd, mstr.Emp_Name, isnull(sum(det.num_dy), 0) as num_day
from dbo.Employee_Mstr mstr
left join dbo.Employee_Nums det on (mstr.Emp_cd = det.Emp_cd)
group by mstr.Emp_cd, mstr.Emp_Name;

should do the job.

Here it's results of SQL execution:

Emp_cd      Emp_Name
----------- --------------------------------------------------
1           A
2           B
3           C

(3 row(s) affected)

Emp_cd      num_dy
----------- -----------
3           4
3           2

(2 row(s) affected)

Emp_cd      Emp_Name                                           num_day
----------- -------------------------------------------------- -----------
1           A                                                  0
2           B                                                  0
3           C                                                  6
Warning: Null value is eliminated by an aggregate or other SET operation.

(3 row(s) affected)

Answer 5

Try this:

select t1.cd, t1.name, sum(t2.num_dy) from employee_Mstr t1 
left join EmployeeLeaveRequest_mstr t2 on t1.cd = t2.cd group by t1.cd, t1.name;