1

ここでは、 値を持つSQL FIDDLE と以下のテーブル スキーマを示します。の条件で選び
たい。私が試したクエリは間違った出力を出します。id_emp,fname_emp,TotalProjectskillsid(1,2)

create table employee_emp(
id_emp int identity(1,1),fname_emp varchar(20),lname_emp varchar(20))
 insert into employee_emp values('John','Cena');
 insert into employee_emp values('Michel','shawn');
 insert into employee_emp values('Jay','mac');
 insert into employee_emp values('David','jackson');

create table skills_skl(
idemp_skl int,idskill_skl int
)
insert into skills_skl values(1,1);
insert into skills_skl values(1,2);
insert into skills_skl values(1,3);
insert into skills_skl values(1,4);
insert into skills_skl values(2,2);
insert into skills_skl values(2,3);
insert into skills_skl values(3,1);
insert into skills_skl values(3,4);
insert into skills_skl values(4,2);

create table employee_ejp(
id_ejp int identity(1,1),idemp_ejp int ,idproject_ejp int)
insert into employee_ejp values(1,1);
insert into employee_ejp values(1,2);
insert into employee_ejp values(1,3);
insert into employee_ejp values(2,3);
insert into employee_ejp values(2,2);
insert into employee_ejp values(3,1);
insert into employee_ejp values(4,4);

試したクエリ

1)

select a.id_emp,a.fname_emp,count(b.id_ejp) TotalP from employee_emp a
join employee_ejp b on a.id_emp=b.idemp_ejp 
group by a.id_emp,a.fname_emp

2)

select a.id_emp,a.fname_emp,count(b.id_ejp) TotalP from employee_emp a
join employee_ejp b on a.id_emp=b.idemp_ejp join skills_skl c on c.idskill_skl=a.id_emp
where c.idskill_skl in (1,2)
group by a.id_emp,a.fname_emp
4

3 に答える 3

1

ここで簡単なアプローチ：

;with emps_1_2 as (
select distinct a.id_emp
  from employee_emp a
  join skills_skl c on c.idemp_skl=a.id_emp     --Be carefull here!
  where c.idskill_skl in (1,2)
)  
select emp.id_emp,
       emp.fname_emp,
       coalesce( COUNT(ejp.id_ejp), 0) as TotalProject 
from       employee_emp emp 
inner join emps_1_2 emp12 
        on emp.id_emp = emp12.id_emp
 left join employee_ejp ejp 
        on emp.id_emp=ejp.idemp_ejp
group by emp.id_emp,emp.fname_emp

結果：

| ID_EMP | FNAME_EMP | TOTALPROJECT |
-------------------------------------
|      4 |     David |            1 |
|      3 |       Jay |            1 |
|      1 |      John |            3 |
|      2 |    Michel |            2 |
于 2013-01-21T10:35:49.777 に答える
1 
select distinct a.id_emp, COUNT(idproject_ejp) over(partition by a.id_emp) 
from #employee_emp a join #skills_skl b 
on a.id_emp = b.idemp_skl
join #employee_ejp c 
on a.id_emp = c.idemp_ejp
where b.idskill_skl in (1,2)
group by a.id_emp,idproject_ejp
于 2013-01-21T10:42:24.510 に答える
0

これでうまくいくはずです：

select emp.id_emp, emp.fname_emp, count(proj.idemp_ejp) as project_count  
from employee_emp emp  
join skills_skl sk  
on emp.id_emp=sk.idemp_skl and sk.idskill_skl in (1,2)  
join employee_ejp proj  
on proj.id_ejp=emp.id_emp  
group by emp.fname_emp, emp.id_emp

最初に従業員テーブルをベースとして作成し、次にスキルを結合句のフィルターとして使用して、データサイズをその場で削減します。
次に、プロジェクト テーブルを結合して、いいえをカウントできるようにします。従業員がテーブルに登場した回数。

短くて甘い。

于 2013-01-21T10:57:03.240 に答える