postgreSQL - 按唯一 ID 计算小时数等

Question

我试图弄清楚如何通过唯一 ID 求和时间（意味着每个 ID 只有一个）。这是一些数据的标记。我需要按 f_name、l_name 和区域分组。我还需要组计数（唯一的 id 计数）和组参与者（只是简单的 id 计数）。

+----+------+--------+------+------+--+ |身份证 |时间 |名称 |姓名 |面积 | | +----+------+--------+------+------+--+ | 1 | 3:30 |杰夫 |选择 |洛杉矶 | | | 1 | 3:30 |杰夫 |选择 |洛杉矶 | | | 1 | 3:30 |杰夫 |选择 |洛杉矶 | | | 2 | 4:00 |杰夫 |选择 |洛杉矶 | | +----+------+--------+------+------+--+

数据应如下所示：

+------+-------+--------+------+-------------+-- ---------------+ |时间 |名称 |南 |面积 |组数 |团体参与者 | +------+-------+--------+------+-------------+-- ---------------+ | 7:30 |杰夫 |选择 |洛杉矶 | 2 | 4 | | | | | | | | +------+-------+--------+------+-------------+-- ---------------+

奖励：如果您可以将 3:30 转换为 3.5

Answer 1

SQL DEMO

WITH cte as (
    SELECT fname, lname, Area,
           COUNT(DISTINCT id) as "Group Count",
           COUNT(id) as "Group Participants", 
           SUM(DISTINCT "time"::time) as "Time"
    FROM Table1
    GROUP BY fname, lname, Area, id 
)    
SELECT fname, lname, Area,
       SUM("Group Count") as "Group Count",
       SUM("Group Participants") as "Group Participants", 
       SUM("Time"::time) as "Time"
FROM cte
GROUP BY fname, lname, Area;

输出