导师严选2025 AI论文网站TOP9:本科生毕业论文必备测评
2026/1/2 22:06:41
表Department:
+---------------+---------+ | Column Name | Type | +---------------+---------+ | id | int | | revenue | int | | month | varchar | +---------------+---------+ 在 SQL 中,(id, month) 是表的联合主键。 这个表格有关于每个部门每月收入的信息。 月份(month)可以取下列值 ["Jan","Feb","Mar","Apr","May","Jun","Jul","Aug","Sep","Oct","Nov","Dec"]。
重新格式化表格,使得每个月都有一个部门 id 列和一个收入列。
以任意顺序返回结果表。
结果格式如以下示例所示。
示例 1:
输入:Department table: +------+---------+-------+ | id | revenue | month | +------+---------+-------+ | 1 | 8000 | Jan | | 2 | 9000 | Jan | | 3 | 10000 | Feb | | 1 | 7000 | Feb | | 1 | 6000 | Mar | +------+---------+-------+输出:+------+-------------+-------------+-------------+-----+-------------+ | id | Jan_Revenue | Feb_Revenue | Mar_Revenue | ... | Dec_Revenue | +------+-------------+-------------+-------------+-----+-------------+ | 1 | 8000 | 7000 | 6000 | ... | null | | 2 | 9000 | null | null | ... | null | | 3 | null | 10000 | null | ... | null | +------+-------------+-------------+-------------+-----+-------------+解释:四月到十二月的收入为空。 请注意,结果表共有 13 列(1 列用于部门 ID,其余 12 列用于各个月份)。
思路:
1,本题主要为列转行
2,可以使用case when 或者poivt
3,部门id分组,sum(每个月的金额)
代码:
select id, sum(case when month = 'Jan' then revenue else null end) Jan_Revenue, sum(case when month = 'Feb' then revenue else null end) Feb_Revenue, sum(case when month = 'Mar' then revenue else null end) Mar_Revenue, sum(case when month = 'Apr' then revenue else null end) Apr_Revenue, sum(case when month = 'May' then revenue else null end) May_Revenue, sum(case when month = 'Jun' then revenue else null end) Jun_Revenue, sum(case when month = 'Jul' then revenue else null end) Jul_Revenue, sum(case when month = 'Aug' then revenue else null end) Aug_Revenue, sum(case when month = 'Sep' then revenue else null end) Sep_Revenue, sum(case when month = 'Oct' then revenue else null end) Oct_Revenue, sum(case when month = 'Nov' then revenue else null end) Nov_Revenue, sum(case when month = 'Dec' then revenue else null end) Dec_Revenue from Department group by id