Question

我正在尝试在空白记录之间填充日期，但是没有成功。试图做多个选择方法，试图加入，但似乎我错过了重点。我想生成缺少日期的记录，从此代码块生成图表。首先，我希望“手动”填写日期，之后，我将重新组织这段代码并将该方法替换为参数。

有人可以帮我这个表情吗？

SELECT
    LOG_LAST AS "data",
    SUM(run_cnt) AS "Number of runs"

FROM

dual l
LEFT OUTER JOIN "LOG_STAT" stat ON
stat."LOG_LAST" = l."CLASS"

WHERE

new_class = '$arg[klasa]'



--SELECT to_date(TRUNC (SYSDATE - ROWNUM), 'DD-MM-YYYY'),
--0
 --FROM dual CONNECT BY ROWNUM < 366

GROUP BY
    LOG_LAST
ORDER BY

LOG_LAST

//编辑：

LOG_LAST只是一个带有日期的列（例如：25.04.2018 15:44:21），run_cnt是一个只有简单数字的列，LOG_STAT是一个包含LOG_LAST和run_cnt的表，new_class是一个带有名称的列我想列出的记录，即使它们不存在。例如：我有一条记录，记录的日期为24-09-2018、23-09-2018、20-09-2018、18-09-2018，并且我想列出没有名称和run_cnt的记录，但会丢失某个时期的日期

Answer 1

尝试填充isull：

SELECT
case when   trim(LOG_LAST) is null then '01-01-2018' 
 else isnull(LOG_LAST,'01-01-2018')end AS data,
    SUM(isnull(run_cnt,0)) AS "Number of runs"

FROM

dual l
LEFT OUTER JOIN "LOG_STAT" stat ON
stat."LOG_LAST" = l."CLASS"

WHERE

new_class = '$arg[klasa]'



--SELECT to_date(TRUNC (SYSDATE - ROWNUM), 'DD-MM-YYYY'),
--0
 --FROM dual CONNECT BY ROWNUM < 366

GROUP BY
    LOG_LAST
ORDER BY

LOG_LAST

Answer 2

它称为数据密集化。来自oracle doc Data Densification for Reporting的示例data densification

with ls as
(
  select trunc(created) as day,object_type new_class, sum(1)  as total
  from user_objects  
  group by trunc(created),object_type
)
, all_dates(day) as
(
  select min(day) from ls
  union all
  select day + 1 from all_dates where day < (select max(day) from ls)
)
select d.day, nvl(ls.total,0),new_class
from all_dates d 
left join ls partition by (ls.new_class)  on ls.day = d.day
order by d.day;

Answer 3

您想要的东西或多或少：

select d.day, sum(ls.run_cnt)
from all_dates d
left join log_stat ls on trunc(ls.log_last) = d.day
where ls.new_class = :klasa
group by d.day
order by d.day;

上面查询中的all_dates表应该包含所有以最小klasa log_last日期开始，以最大klasa log_last日期结束的日期。您可以通过递归查询获得这些日期。

with ls as
(
  select trunc(log_last) as day, sum(run_cnt) as total
  from log_stat
  where new_class = :klasa
  group by trunc(log_last)
)
, all_dates(day) as
(
  select min(day) from ls
  union all
  select day + 1 from all_dates where day < (select max(day) from ls)
)
select d.day, ls.total
from all_dates d
left join ls on ls.day = d.day
order by d.day;

如何填写空白记录之间的缺失日期？

3 个答案: