获取距离过去和未来最近的实例

Question

我有一张包含appointment记录的表，并且用户选择日期范围（begin_date,end_date）。我希望获得落在此日期范围内的appointment以及过去和未来在此日期范围之外（又名，上一次出现和下一次出现）的最接近的实例。

我认为最好的方法是使用CTE和自联接，但我接受另一种策略。这是我当前的查询：

WITH present AS 
(SELECT appt.ewssubject, appt.ewsstart::DATE, appt.ewsend::DATE 
FROM appointment appt 
WHERE (appt.ewsstart, appt.ewsend) OVERLAPS (begin_date::DATE, end_date::DATE)) 

SELECT 
present.ewssubject, present.ewsstart, present.ewsend, 
past.ewssubject AS pastsubject, past.ewsstart::DATE AS paststart,past.ewsend::DATE AS pastend, 
future.ewssubject AS futuresubject, future.ewsstart::date AS futurestart, future.ewsend::date AS futureend 

FROM present 
LEFT JOIN appointment AS past USING (ewssubject) 
LEFT JOIN appointment AS future USING (ewssubject) 

WHERE 
present.ewssubject = past.ewssubject AND 
present.ewssubject = future.ewssubject AND 
past.ewsend < present.ewsstart AND 
future.ewsstart > present.ewsend 

ORDER BY present.ewsstart ASC 

我得到了巨大的appointments列表，有很多重复的 - 像这样：

subject start  end   last_start last_end next_start next_end 
DINNER 2015-09-18 2015-09 18 2015-09-17 2015-09-17 2015-09-19 2015-09-19 
DINNER 2015-09-18 2015-09 18 2015-09-17 2015-09-17 2015-09-19 2015-09-19 
... // more repeats! :(

所有我想要做的就是减少重复数量，如下格式：

subject start  end   last_start last_end next_start next_end 
DINNER 2015-09-18 2015-09-18 2015-09-17 2015-09-17 2015-09-19 2015-09-19 
DINNER 2015-09-21 2015-09-21 2015-09-18 2015-09-18 2015-10-02 2015-10-02 
... // and so on 

nb一个appointment可以跨越多天。

我该如何解决我的问题？或者还有另一个我可以写的？

Answer 1

你没有输入详细了解您的数据，所以我不知道这是否是一个好主意，但你也许可以用window functions：

select 
    ewssubject, ewsstart, ewsend, 
    lag(ewsstart) over (partition by ewssubject order by ewstart) prior_start, 
    lag(ewsend) over (partition by ewssubject order by ewstart) prior_end, 
    lead(ewsstart) over (partition by ewssubject order by ewstart) next_start, 
    lead(ewsend) over (partition by ewssubject order by ewstart) next_end 
from appointment 
order by ewstart;