我在下面有一个SQL查询
select
row_number()
OVER (ORDER BY hapf.position_code) Identifier,
hapf.position_code as position_code,
pg.name as gradename,
pgsfv.name as stepname
from
hr_all_positions_f hapf, PER_VALID_GRADES_F pvgf, per_grades pg, PER_GRADE_STEPS_F_VL pgsfv
where
hapf.position_id = pvgf.position_id
and pvgf.grade_id = pg.grade_id
and pgsfv.grade_id = pg.grade_id
and hapf.position_code = 'PRGUSPOS084'
and TO_CHAR(hapf.effective_end_date, 'YYYY-MM-DD') = '4712-12-31'
输出低于
Identifier Position_Code Gradename Stepname
1 PRGUSPOS084 Salary05 Step01
2 PRGUSPOS084 Salary05 Step02
3 PRGUSPOS084 Salary05 Step03
4 PRGUSPOS084 Salary06 Step01
5 PRGUSPOS084 Salary06 Step02
6 PRGUSPOS084 Salary06 Step03
7 PRGUSPOS084 Salary07 Step01
8 PRGUSPOS084 Salary07 Step02
9 PRGUSPOS084 Salary07 Step03
我已经使用case来将GradeName列转换为如下所示的行
SQL
select
row_number()
OVER (ORDER BY position_code) RN,
position_code as pos,
--stepname as step,
max(case when Identifier = 1 then gradename end) as Grade1,
max(case when Identifier = 2 then gradename end) as Grade2,
max(case when Identifier = 3 then gradename end) as Grade3
from
(
select
row_number()
OVER (ORDER BY hapf.position_code) Identifier,
hapf.position_code as position_code,
pg.name as gradename,
pgsfv.name as stepname
from
hr_all_positions_f hapf, PER_VALID_GRADES_F pvgf, per_grades pg, PER_GRADE_STEPS_F_VL pgsfv
where
hapf.position_id = pvgf.position_id
and pvgf.grade_id = pg.grade_id
and pgsfv.grade_id = pg.grade_id
and hapf.position_code = 'PRGUSPOS084'
and TO_CHAR(hapf.effective_end_date, 'YYYY-MM-DD') = '4712-12-31'
) x
group by position_code
此查询的输出如下
RN POS GRADE1 GRADE2 GRADE3
1 PRGUSPOS084 Salary 05 Salary 06 Salary 07
但我想要输出如下
NO POS GRADE1 GRADE2 GRADE3 Grade1Step1 Grade1Step2 Grade1Step3 Grade2Step1 Grade2Step2 Grade2Step3 Grade3Step1 Grade3Step2 Grade3Step3
1 PRGUSPOS084 Salary 05 Salary 06 Salary 07 Step01 Step02 Step03 Step01 Step02 Step03 Step01 Step02 Step03
如何使用另一个案例陈述?
我尝试了以下但输出不正确,请帮忙。
select
pos,
Grade1,
Grade2,
Grade3,
max(case when RN = 1 then step end) as stepname1,
max(case when RN = 2 then step end) as stepname2,
max(case when RN = 3 then step end) as stepname3
from
(
select
row_number()
OVER (ORDER BY position_code) RN,
position_code as pos,
stepname as step,
max(case when Identifier = 1 then gradename end) as Grade1,
max(case when Identifier = 2 then gradename end) as Grade2,
max(case when Identifier = 3 then gradename end) as Grade3
from
(
select
row_number()
OVER (ORDER BY hapf.position_code) Identifier,
hapf.position_code as position_code,
pg.name as gradename,
pgsfv.name as stepname
from
hr_all_positions_f hapf, PER_VALID_GRADES_F pvgf, per_grades pg, PER_GRADE_STEPS_F_VL pgsfv
where
hapf.position_id = pvgf.position_id
and pvgf.grade_id = pg.grade_id
and pgsfv.grade_id = pg.grade_id
and hapf.position_code = 'PRGUSPOS084'
and TO_CHAR(hapf.effective_end_date, 'YYYY-MM-DD') = '4712-12-31'
) x
group by position_code
, stepname
) z group by pos, Grade1, Grade2, Grade3
您可以使用不同的partition-by clauses调用内部查询中的多个分析函数,而不只是一个;然后从那些工作 - 梳理案例表达中的成绩和步骤的分析排名。
因为我没有你的基表我已经从你的样本中给出了固定值作为进一步的内部查询来演示这个想法,只需要一个额外的值来显示第二行输出:
select position_num,
position_code,
max(case when grade_num = 1 then grade_name end) as grade1,
max(case when grade_num = 2 then grade_name end) as grade2,
max(case when grade_num = 3 then grade_name end) as grade3,
max(case when grade_num = 1 and grade_step_num = 1 then step_name end) as Grade1Step1,
max(case when grade_num = 1 and grade_step_num = 2 then step_name end) as Grade1Step2,
max(case when grade_num = 1 and grade_step_num = 3 then step_name end) as Grade1Step3,
max(case when grade_num = 2 and grade_step_num = 1 then step_name end) as Grade2Step1,
max(case when grade_num = 2 and grade_step_num = 2 then step_name end) as Grade2Step2,
max(case when grade_num = 2 and grade_step_num = 3 then step_name end) as Grade2Step3,
max(case when grade_num = 3 and grade_step_num = 1 then step_name end) as Grade3Step1,
max(case when grade_num = 3 and grade_step_num = 2 then step_name end) as Grade3Step2,
max(case when grade_num = 3 and grade_step_num = 3 then step_name end) as Grade3Step3
from (
select
dense_rank() over (order by position_code) position_num,
position_code,
grade_name,
step_name,
dense_rank() over (partition by position_code order by grade_name) as grade_num,
dense_rank() over (partition by position_code, grade_name order by step_name) as grade_step_num
from
(
select 'PRGUSPOS084' as position_code, 'Salary05' as grade_name, 'Step01' as step_name from dual
union all select 'PRGUSPOS084', 'Salary05', 'Step02' from dual
union all select 'PRGUSPOS084', 'Salary05', 'Step03' from dual
union all select 'PRGUSPOS084', 'Salary06', 'Step01' from dual
union all select 'PRGUSPOS084', 'Salary06', 'Step02' from dual
union all select 'PRGUSPOS084', 'Salary06', 'Step03' from dual
union all select 'PRGUSPOS084', 'Salary07', 'Step01' from dual
union all select 'PRGUSPOS084', 'Salary07', 'Step02' from dual
union all select 'PRGUSPOS084', 'Salary07', 'Step03' from dual
union all select 'PRGUSPOS085', 'Salary06', 'Step02' from dual
)
)
group by position_num, position_code;
获得:
POSITION_NUM POSITION_CO GRADE1 GRADE2 GRADE3 GRADE1 GRADE1 GRADE1 GRADE2 GRADE2 GRADE2 GRADE3 GRADE3 GRADE3
------------ ----------- -------- -------- -------- ------ ------ ------ ------ ------ ------ ------ ------ ------
1 PRGUSPOS084 Salary05 Salary06 Salary07 Step01 Step02 Step03 Step01 Step02 Step03 Step01 Step02 Step03
2 PRGUSPOS085 Salary06 Step02
所以你真正的查询可能最终会像:
select position_num,
position_code,
max(case when grade_num = 1 then grade_name end) as grade1,
max(case when grade_num = 2 then grade_name end) as grade2,
max(case when grade_num = 3 then grade_name end) as grade3,
max(case when grade_num = 1 and grade_step_num = 1 then step_name end) as Grade1Step1,
max(case when grade_num = 1 and grade_step_num = 2 then step_name end) as Grade1Step2,
max(case when grade_num = 1 and grade_step_num = 3 then step_name end) as Grade1Step3,
max(case when grade_num = 2 and grade_step_num = 1 then step_name end) as Grade2Step1,
max(case when grade_num = 2 and grade_step_num = 2 then step_name end) as Grade2Step2,
max(case when grade_num = 2 and grade_step_num = 3 then step_name end) as Grade2Step3,
max(case when grade_num = 3 and grade_step_num = 1 then step_name end) as Grade3Step1,
max(case when grade_num = 3 and grade_step_num = 2 then step_name end) as Grade3Step2,
max(case when grade_num = 3 and grade_step_num = 3 then step_name end) as Grade3Step3
from (
select
dense_rank() over (order by hapf.position_code) position_num,
hapf.position_code,
pg.name as grade_name,
pgsfv.name as step_name,
dense_rank() over (partition by hapf.position_code order by pg.name) as grade_num,
dense_rank() over (partition by hapf.position_code, pg.name order by pgsfv.name)
as grade_step_num
from
hr_all_positions_f hapf
join per_valid_grades_f pvgf on hapf.position_id = pvgf.position_id
join per_grades pg on pvgf.grade_id = pg.grade_id
join per_grade_steps_f_vl pgsfv on pgsfv.grade_id = pg.grade_id
where hapf.position_code = 'PRGUSPOS084'
and hapf.effective_end_date = date '4712-12-31'
)
group by position_num, position_code;
我冒昧地将你的内部查询切换为使用现代连接语法;并将你的effective_end_date
与实际日期进行比较,而不是将其转换为字符串 - 这通常是一个坏主意,因为它会阻止使用该列的索引。