如何用SQL计算行之间的差异？

Question

我有一个任务是使用一个新列创建一个表，其中包含当前行和下一行中的年龄之间的差异。 年龄应按降序排列。 是否可以使用SQL？ 我不确切知道我应该使用什么sql，而且在任务中没有指定。

我明白括号内应该有东西

SELECT name, age, (...) AS difference
FROM Animals
ORDER BY age DESC;

我有“动物”这个表

id  |  name  | age
 1  | Molly  |  4
 2  | Jackson|  8 
 3  | Wonka  |  38
 4  | Polly  |  7

结果表应如下所示：

  name  | age | difference
 Wonka  | 38  |  30
 Jackson|  8  |  1
 Polly  |  7  |  3
 Molly  |  4  |

Answer 1

你需要lead()函数：

SELECT 
  name, 
  age, 
  age - lead(age) over (order by age desc, name) AS difference
FROM Animals
ORDER BY age DESC

看演示 。
结果：

> name    | age | difference
> :------ | --: | ---------:
> Wonka   |  38 |         30
> Jackson |   8 |          1
> Polly   |   7 |          3
> Molly   |   4 |       

Answer 2

您将使用lag() / lead()并按order by ：

select a.*,
       (a - lag(age) over (order by age)) as diff
from animals a
order by age desc;

的order by的lag()不需要的匹配order by在外部查询。 第一个定义“前一行”。 第二个是数据的呈现。

Answer 3

您可以使用MAX函数实现所需的结果，该函数在相应的行窗口上运行。

with Animals as (
  select 1 as id, 'Molly' as name, 4 as age union all
  select 2, 'Jackson', 8 union all
  select 3, 'Wonka', 38 union all
  select 4, 'Polly', 7
)
select
  name, age,
  age - max(age) over(
          order by age
          rows between unbounded preceding
          and 1 preceding
        )
from Animals
order by age desc;

输出：

+---------+-----+------------+
|  name   | age | difference |
+---------+-----+------------+
| Wonka   |  38 | 30         |
| Jackson |   8 | 1          |
| Polly   |   7 | 3          |
| Molly   |   4 | NULL       |
+---------+-----+------------+