Question

我有一个包含以下列的表

table: route
columns: id, location, order_id

并且它具有诸如

之类的值

id, location, order_id
1, London, 12
2, Amsterdam, 102
3, Berlin, 90
5, Paris, 19

是否可以在postgres中执行一个sql select语句，它将返回每一行以及带有下一个最高order_id的id？所以我想要像...这样的东西。

id, location, order_id, next_id
1, London, 12, 5
2, Amsterdam, 102, NULL
3, Berlin, 90, 2
5, Paris, 19, 3

由于

Answer 1

select 
  id, 
  location, 
  order_id,
  lag(id) over (order by order_id desc) as next_id
from your_table

Answer 2

首先创建测试平台：

CREATE TABLE route (id int4, location varchar(20), order_id int4);
INSERT INTO route VALUES
    (1,'London',12),(2,'Amsterdam',102),
    (3,'Berlin',90),(5,'Paris',19);

查询：

WITH ranked AS (
    SELECT id,location,order_id,rank() OVER (ORDER BY order_id)
      FROM route)
SELECT b.id, b.location, b.order_id, n.id
  FROM ranked b
  LEFT JOIN ranked n ON b.rank+1=n.rank
  ORDER BY b.id;

您可以在documentation。

中阅读有关窗口功能的更多信息

Answer 3

是：

select * ,
(select top 1 id from routes_table where order_id > main.order_id order by 1 desc)
from routes_table main

在sql中创建一个伪链表

3 个答案: