Question

我有这样的章节表：

id | title    | sort_number | book_id
1  | 'Chap 1' | 3           | 1
5  | 'Chap 2' | 6           | 1
8  | 'About ' | 1           | 1
9  | 'Chap 3' | 9           | 1
10 | 'Attack' | 1           | 2

Id是唯一的，sort_number对于同一本书（book_id）是唯一的

1）如果我只有当前章节ID，如何按sort_number排序的3个章节（当前，下一个和上一个）的所有数据（3行）如何加载？

2）我如何加载当前章节数据（1行），并且只有下一个ID或上一个ID（如果存在）？

Answer 1

可以使用window functions

select id, title, sort_number, book_id, 
       lag(id) over w as prev_chapter,
       lead(id) over w as next_chapter
from chapters
window w as (partition by book_id order by sort_number);

您的示例数据返回：

id | title  | sort_number | book_id | prev_chapter | next_chapter
---+--------+-------------+---------+--------------+-------------
 8 | About  |           1 |       1 |              |            1
 1 | Chap 1 |           3 |       1 |            8 |            5
 5 | Chap 2 |           6 |       1 |            1 |            9
 9 | Chap 3 |           9 |       1 |            5 |             
10 | Attack |           1 |       2 |              |

上面的查询现在可以用来回答您的两个问题：

1）

select id, title, sort_number, book_id
from (
  select id, title, sort_number, book_id, 
         --first_value(id) over w as first_chapter,
         lag(id) over w as prev_chapter_id,
         lead(id) over w as next_chapter_id
  from chapters
  window w as (partition by book_id order by sort_number)
) t
where 1 in (id, prev_chapter_id, next_chapter_id)

2）

select *
from (
  select id, title, sort_number, book_id, 
         lag(id) over w as prev_chapter_id,
         lead(id) over w as next_chapter_id
  from chapters
  window w as (partition by book_id order by sort_number)
) t
where id = 1

按ID选择行，它是按某个值排序的最接近的行。 PostgreSQL的

1 个答案: