差距和岛屿 - 使用 Postgresql 获取某个日期范围内的失业日期列表答案

【问题标题】：Gaps and Islands - get a list of dates unemployed over a date range with Postgresl差距和岛屿 - 使用 Postgresql 获取某个日期范围内的失业日期列表
【发布时间】：2019-07-08 17:02:09
【问题描述】：

我有一个名为 Position 的表，在这个表中，我有以下内容，包括日期 (yyyy-mm-dd)，下面是就业日期的简化视图

id, person_id, start_date, end_date  , title
1 , 1        , 2001-12-01, 2002-01-31, 'admin'
2 , 1        , 2002-02-11, 2002-03-31, 'admin'
3 , 1        , 2002-02-15, 2002-05-31, 'sales'
4 , 1        , 2002-06-15, 2002-12-31, 'ops'

我希望能够计算就业差距，假设某些日期重叠以为 id=1 的人生成以下输出

person_id, start_date, end_date  , last_position_id, gap_in_days
1        , 2002-02-01, 2002-02-10, 1               , 10
1        , 2002-06-01, 2002-06-14, 3               , 14

我查看了许多解决方案、UNIONS、物化视图、生成日历日期范围的表等。我真的不确定什么是最好的方法。是否有一个查询可以让我完成这项工作？

【问题讨论】：

你是什么意思assuming some of the dates overlap？
什么是last_position_id？
一个人可以有两个职位，因此日期重叠
last_position_id 是间隙前的位置的id

标签： postgresql date-range gaps-and-islands

【解决方案1】：

首先你需要找到重叠的日期Determine Whether Two Date Ranges Overlap
然后将这些范围合并为一个范围并保留最后一个 id
最后计算一个end_date和下一个start_date - 1之间的天数范围

SQL DEMO

with find_overlap as (
  SELECT t1."id" as t1_id, t1."person_id", t1."start_date", t1."end_date",
         t2."id" as t2_id, t2."start_date" as t2_start_date, t2."end_date" as t2_end_date
  FROM Table1 t1
  LEFT JOIN Table1 t2
    ON t1."person_id" = t2."person_id"
   AND t1."start_date" <= t2."end_date"
   AND t1."end_date"   >= t2."start_date"
   AND t1.id < t2.id
), merge_overlap as (
  SELECT 
         person_id,
         start_date,
         COALESCE(t2_end_date, end_date) as  end_date,
         COALESCE(t2_id, t1_id) as last_position_id
  FROM find_overlap
  WHERE t1_id NOT IN (SELECT t2_id FROM find_overlap WHERE t2_ID IS NOT NULL)
), cte as (
  SELECT *, 
         LEAD(start_date) OVER (partition by person_id order by start_date) next_start
  FROM merge_overlap 
) 
SELECT *, 
       DATE_PART('day', 
                  (next_start::timestamp - INTERVAL '1 DAY') - end_date::timestamp
                ) as days 
FROM cte
WHERE next_start IS NOT NULL

输出

| person_id | start_date |   end_date | last_position_id | next_start | days |
|-----------|------------|------------|------------------|------------|------|
|         1 | 2001-12-01 | 2002-01-31 |                1 | 2002-02-11 |   10 |
|         1 | 2002-02-11 | 2002-05-31 |                3 | 2002-06-15 |   14 |

【讨论】：

这假设只有两个日期之间的简单重叠。如果您有三向重叠，可能需要先进行递归查询
为什么我得到了正确的答案。即使我的答案是正确的，S-Man 的答案似乎更简单

【解决方案2】：

step-by-step demo:db<>fiddle

您只需要lead() window function。有了这个，您可以获取当前行的值（在这种情况下为start_date）。

SELECT
    person_id,
    end_date + 1 AS start_date,
    lead - 1 AS end_date,
    id AS last_position_id,
    lead - (end_date + 1) AS gap_in_days
FROM (
    SELECT 
        *,
        lead(start_date) OVER (PARTITION BY person_id ORDER BY start_date)
    FROM
        positions
) s
WHERE lead - (end_date + 1) > 0

获得下一个start_date 后，您可以将其与当前的end_date 进行比较。如果它们不同，那么您就有差距。这些正值可以在WHERE 子句中过滤。

（如果2个位置重叠，则diff为负数，可以忽略。）

【讨论】：