题目
表: Calls
+-------------+---------+ | Column Name | Type | +-------------+---------+ | from_id | int | | to_id | int | | duration | int | +-------------+---------+ 该表没有主键,可能存在重复项。 该表包含 from_id 与 to_id 间的一次电话的时长。 from_id != to_id
编写 SQL 语句,查询每一对用户 (person1, person2) 之间的通话次数和通话总时长,其中 person1 < person2 。
以 任意顺序 返回结果表。
查询结果格式如下示例所示。
示例 1:
输入: Calls 表: +---------+-------+----------+ | from_id | to_id | duration | +---------+-------+----------+ | 1 | 2 | 59 | | 2 | 1 | 11 | | 1 | 3 | 20 | | 3 | 4 | 100 | | 3 | 4 | 200 | | 3 | 4 | 200 | | 4 | 3 | 499 | +---------+-------+----------+ 输出: +---------+---------+------------+----------------+ | person1 | person2 | call_count | total_duration | +---------+---------+------------+----------------+ | 1 | 2 | 2 | 70 | | 1 | 3 | 1 | 20 | | 3 | 4 | 4 | 999 | +---------+---------+------------+----------------+ 解释: 用户 1 和 2 打过 2 次电话,总时长为 70 (59 + 11)。 用户 1 和 3 打过 1 次电话,总时长为 20。 用户 3 和 4 打过 4 次电话,总时长为 999 (100 + 200 + 200 + 499)。
解题
方法一:
通过子查询,先用if保证person1比person2小,然后再去处理
select person1, person2, count(*) as call_count, sum(duration) as total_duration from( select if(from_id>to_id,to_id,from_id) as person1, if(from_id>to_id,from_id,to_id) as person2, duration from Calls )as c group by person1,person2
方法二:
使用了least 和 greatest
select from_id as person1, to_id as person2, count(*) as call_count, sum(duration) as total_duration from Calls group by least(from_id,to_id),greatest(from_id,to_id);