Question

我在Postgres 9.1数据库中有以下表结构，但如果可能，理想的解决方案应该是DB不可知的：

Table: users
|id|username|
|1 |one     |
|2 |two     |
|3 |three   |

Table: items
|id|userid|itemname|created  |
|1 |1     |a       |timestamp|
|2 |1     |b       |timestamp|
|3 |1     |c       |timestamp|
|4 |2     |d       |timestamp|
|5 |2     |e       |timestamp|
|6 |2     |f       |timestamp|
|7 |3     |g       |timestamp|
|8 |3     |h       |timestamp|
|9 |3     |i       |timestamp|

我有一个查询（对于一个视图），它提供了下一个和上一个item.id。

e.g。

View: UserItems
|id|userid|itemname|nextitemid|previtemid|created  |
|1 |1     |a       |2         |null      |timestamp|
|2 |1     |b       |3         |1         |timestamp|
|3 |1     |c       |4         |2         |timestamp|
|4 |2     |d       |5         |3         |timestamp|
|5 |2     |e       |6         |4         |timestamp|
|6 |2     |f       |7         |5         |timestamp|
|7 |3     |g       |8         |6         |timestamp|
|8 |3     |h       |9         |7         |timestamp|
|9 |3     |i       |null      |8         |timestamp|

我可以使用以下查询执行此操作：

SELECT
  DISTINCT i.id AS id,
  i.userid AS userid,
  i.itemname AS itemname,
  LEAD(i.id) OVER (ORDER BY i.created DESC) AS nextitemid,
  LAG(i.id) OVER (ORDER BY i.created DESC) AS previtemid,
  i.created AS created
FROM items i
  LEFT JOIN users u
  ON i.userid = u.id
ORDER BY i.created DESC;

您能帮忙解决以下问题：

1）是否有办法使ids包裹，即

nextitemid列的最后一行中的NULL itemid应为1
previtemid列第一行中的NULL itemid应为9

2）是否有一种高效的方法来按用户ID分组下一个和之前的itemid。

NB：在此示例中，用户的itemid是顺序的，实际数据不是这种情况，每个用户的itemid是交错的。

View: UserItems
|id|userid|itemname|nextitemid|previtemid|nextuseritemid|prevuseritemid|created  |
|1 |1     |a       |2         |9         |2             |3             |timestamp|
|2 |1     |b       |3         |1         |3             |1             |timestamp|
|3 |1     |c       |4         |2         |1             |2             |timestamp|
|4 |2     |d       |5         |3         |5             |6             |timestamp|
|5 |2     |e       |6         |4         |6             |4             |timestamp|
|6 |2     |f       |7         |5         |4             |5             |timestamp|
|7 |3     |g       |8         |6         |8             |9             |timestamp|
|8 |3     |h       |9         |7         |9             |7             |timestamp|
|9 |3     |i       |1         |8         |7             |8             |timestamp|

Answer 1

Q1：FIRST_VALUE / LAST_VALUE

Q2：PARTITION BY（正如Roman Pekar已经建议的那样）

<强> SEE FIDDLE HERE

SELECT
  DISTINCT i.id AS id,
  i.userid AS userid,
  i.itemname AS itemname,
  COALESCE(LEAD(i.id)        OVER (ORDER BY i.created DESC)
          ,FIRST_VALUE(i.id) OVER (ORDER BY i.created DESC ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING)) AS nextitemid,
  COALESCE(LAG(i.id)         OVER (ORDER BY i.created DESC)
          ,LAST_VALUE(i.id)  OVER (ORDER BY i.created DESC ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING)) AS previtemid,
  COALESCE(LEAD(i.id)        OVER (PARTITION BY i.userid ORDER BY i.created DESC)
          ,FIRST_VALUE(i.id) OVER (PARTITION BY i.userid ORDER BY i.created DESC ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING)) AS nextuseritemid,
  COALESCE(LAG(i.id)         OVER (PARTITION BY i.userid ORDER BY i.created DESC)
          ,LAST_VALUE(i.id)  OVER (PARTITION BY i.userid ORDER BY i.created DESC ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING)) AS prevuseritemid,
  i.created AS created
FROM items i
  LEFT JOIN users u
  ON i.userid = u.id
ORDER BY i.created DESC;

Answer 2

更新我忘记了PostgreSQL中的first_value and last_value functions，感谢他提醒我的不好意思。但是，他的查询不起作用，因为last_value正在使用默认窗口RANGE BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW并且不会返回正确的结果，因此您必须更改over子句中的范围或使用first_value order by asc：

select
    i.id as id,
    i.userid as userid,
    i.itemname as itemname,
    coalesce(
        lead(i.id) over(order by i.created desc),
        first_value(i.id) over(order by i.created desc)
    ) as nextitemid,
    coalesce(
        lag(i.id) over(order by i.created desc),
        first_value(i.id) over(order by i.created asc)
    ) as previtemid,
    coalesce(
        lead(i.id) over(partition by i.userid order by i.created desc),
        first_value(i.id) over(partition by i.userid order by i.created desc)
    ) as nextuseritemid,
    coalesce(
        lag(i.id) over(partition by i.userid order by i.created desc),
        first_value(i.id) over(partition by i.userid order by i.created asc)
    ) as prevuseritemid,
    i.created as created
from items as i
   left outer join users as u on u.id = i.userid
order by i.created desc

sql fiddle demo

以前的版本
我想你可以这样做：

SELECT
  i.id AS id,
  i.userid AS userid,
  i.itemname AS itemname,
  coalesce(
      LEAD(i.id) OVER (ORDER BY i.created DESC),
      (select t.id from items as t order by t.created desc limit 1)
  ) AS nextitemid,
  coalesce(
      LAG(i.id) OVER (ORDER BY i.created DESC),
      (select t.id from items as t order by t.created asc limit 1)
  ) AS previtemid,
  coalesce(
      LEAD(i.id) OVER (partition by i.userid ORDER BY i.created DESC),
      (select t.id from items as t where t.userid = i.userid order by t.created desc limit 1)
  ) AS nextuseritemid,
  coalesce(
      LAG(i.id) OVER (partition by i.userid ORDER BY i.created DESC),
      (select t.id from items as t where t.userid = i.userid order by t.created asc limit 1)
  ) AS prevuseritemid,
  i.created AS created
FROM items i
  LEFT JOIN users u
  ON i.userid = u.id
ORDER BY i.created DESC;

sql fiddle demo

Postgres下一个/上一行SQL查询

2 个答案: