Question

我想做一个sql语句，它在一个表中计算2个Date值的减法。如果值为负，我只想将其显示为0值。

数字值是付款处于当前状态的秒数。我将它用技巧转换为时间值（日期类型）。

我目前的代码是

SELECT 
max(CASE WHEN t1.time_event < SYSDATE and t2.time_event > SYSDATE THEN to_char(to_date(max(round(SYSDATE - t1.time_event) * 24 * 60 * 60)),'ssssss'),'hh24:mi:ss') else to_char(to_date(0)) END) as "current_age"
from tbl_dummyfeed t1 join tbl_dummyfeed t2 on t1.payment_Id = t2.payment_id
where t1.event = 'accepted' and t2.event = 'enriched');

Answer 1

这是有效的，只有转换到时间hh24：mi：ss仍然需要发生。

SELECT max(CASE WHEN t1.time_event < SYSDATE and t2.time_event > SYSDATE THEN round((SYSDATE - t1.time_event) * 24 * 60 * 60) else 0 END) as "current_age"
    from tbl_dummyfeed t1 join tbl_dummyfeed t2 on t1.payment_id= t2.payment_id
    where t1.event = 'accepted' and t2.event = 'enriched';

当我将转换添加到hh24：mm：ss解决方案看起来像这样

SELECT to_char(to_date(max(CASE WHEN t1.time_event < SYSDATE and t2.time_event > SYSDATE THEN round((SYSDATE - t1.time_event) * 24 * 60
* 60) else 0 END),'sssss'),'hh24:mi:ss') as "current_age" from tbl_dummyfeed t1 join tbl_dummyfeed t2 on t1.trax_Id = t2.trax_id where t1.event = 'accepted' and t2.event = 'enriched';

这是我的问题的唯一好方法。希望这有助于人们。

Answer 2

您可以使用类似的日期'技巧'，即将小数天差异添加到名义日期，其中时间部分为午夜 - 您可以使用固定日期或trunc(sysdate)，因为时间结束为午夜 - 而不必乘以24 * 60 * 60。（你的to_date()解决方案隐含地执行相同的操作，有效地将当前月份的第一天的午夜数添加到午夜;但这可能会更清楚一点。但您也可以将case子句移到where过滤器

中

select to_char(date '1970-01-01'
  + nvl(max(sysdate - t1.time_event), 0), 'HH24:MI:SS') as "current_age"
from tbl_dummyfeed t1
join tbl_dummyfeed t2
on t1.trax_id = t2.trax_id
where t1.event = 'accepted'
and t1.time_event < sysdate
and t2.event = 'enriched'
and t2.time_event > sysdate;

您还可以使用分析方法，因此您只需要打一次表，使用子查询将每个“丰富”时间与该ID的先前“已接受”时间配对，然后根据当前过滤时间：

select to_char(date '1970-01-01'
  + nvl(max(sysdate - time_accepted), 0), 'HH24:MI:SS') as "current_age"
from (
  select last_value(case when event = 'accepted' then time_event end ignore nulls)
      over (partition by trax_id order by time_event) as time_accepted,
    case when event = 'enriched' then time_event end as time_enriched
  from tbl_dummyfeed
)
where time_accepted < sysdate
and time_enriched > sysdate;

SQL：结合案例和数字到日期的转换

2 个答案: