Question

我正在运行一些报告查询，我想在4周的特定时间内扩展每条记录的结果。

这是当前查询：

select
  job_id,
  week,
  count(*),
  sum(count(*)) over (partition by job_id)
from candidates
group by job_id, week

当前结果：

 job_id | week | count | sum 
--------+------+-------+------
   3258 |    1 |    21 |  23 
   3258 |    2 |     2 |  23 
   3259 |    1 |     1 |   4 
   3259 |    4 |     1 |   4

但理想情况下，我想扩展4周的特定范围：

所需结果

 job_id | week | count | sum 
--------+------+-------+-----
   3258 |    1 |    21 |  23 
   3258 |    2 |     2 |  23 
   3258 |    3 |     0 |  23 # added row with 0 count
   3258 |    4 |     0 |  23 # added row with 0 count
   3259 |    1 |     1 |   4 
   3259 |    2 |     0 |   4 # added row with 0 count
   3259 |    3 |     0 |   4 # added row with 0 count
   3259 |    4 |     1 |   4

使用左联接不会返回预期的结果，如您在此SQL fiddle

中所见

模式（PostgreSQL v9.6）

CREATE TABLE candidates(
   job_id integer,
   week integer,
   count1 integer,
   sum1 integer
);

INSERT INTO candidates(job_id, week, count1, sum1) VALUES (3984, 1, 13, 26);
INSERT INTO candidates(job_id, week, count1, sum1) VALUES (3984, 2, 13, 26);

INSERT INTO candidates(job_id, week, count1, sum1) VALUES (3985, 1, 42, 46);
INSERT INTO candidates(job_id, week, count1, sum1) VALUES (3985, 4, 3, 46);

查询＃1

select
  c.job_id,
  weeks.week_nr as week,
  c.count1,
  c.sum1
from generate_series(1,4) as weeks(week_nr)
left join candidates c on c.week = weeks.week_nr 
order by c.job_id, week;

| job_id | week | count | sum |
| ------ | ---- | ----- | --- |
| 3984   | 1    | 1     | 2   |
| 3984   | 2    | 1     | 2   |
| 3985   | 1    | 1     | 2   |
| 3985   | 4    | 1     | 2   |
| null   | 3    | null  | null|

Answer 1

在Postgresql中，我们可以使用 generate_series（）（或）其中条件

 select job_id,week,count(*),sum(count(*)) over (partition by job_id)
 from generate_series(1,4) as weeks(week_nr)
 left join candidates c on c.week = weeks.week_nr 
 group by job_id, week order by job_id,week;

                       (or)

 select job_id,week,count(*),sum(count(*)) over (partition by job_id)
 from candidates where week>=1 and week<=4
 group by job_id, week;

跨数组值扩展结果集

1 个答案: