问题
I have the following table structures in a Postgres 9.1 database but the ideal solution should be DB agnostic if possible:
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|
I have a query (for a view) which provides the next and previous 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|
I can do this with the following query:
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;
Can you help to solve the following problems:
1) Is there a way to make the ids wrap i.e.
- the NULL itemid in the last row of the nextitemid column should be 1
- the NULL itemid in the first row of the previtemid column should be 9
2) is there a performant way to group the next and previous itemids by userid e.g.
NB: in this example the itemids for a user are sequential, this is not the case for real data, the itemids for each user are interleaved.
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|
回答1:
Q1: FIRST_VALUE/LAST_VALUE
Q2: PARTITION BY (as Roman Pekar already suggested)
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;
回答2:
update I've forgot about first_value and last_value functions in PostgreSQL, thanks to dnoeth he reminded me about it. However, his query is not working, because last_value is working with default window RANGE BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW and will not return correct results, so you either have to change range inside the over clause or use first_value with 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
previous version
I think you could do this:
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
来源：https://stackoverflow.com/questions/18290572/postgres-next-previous-row-sql-query