列表

详情


SQL190. 某乎问答11月份日人均回答量

描述

现有某乎问答创作者回答情况表answer_tb如下(其中answer_date表示创作日期、author_id指创作者编号、issue_id表示问题id、char_len表示回答字数):
answer_date author_id issue_id char_len
2021-11-01 101 E001 150
2021-11-01
101 E002 200
2021-11-01
102 C003 50
2021-11-01
103 P001 35
2021-11-01
104 C003 120
2021-11-01
105 P001 125
2021-11-01
102 P002 105
2021-11-02
101 P001 201
2021-11-02
110 C002 200
2021-11-02
110 C001 225
2021-11-02
110 C002 220
2021-11-03
101 C002 180
2021-11-04
109 E003 130
2021-11-04
109 E001 123
2021-11-05
108 C001 160
2021-11-05
108 C002 120
2021-11-05
110 P001 180
2021-11-05
106 P002 45
2021-11-05
107 E003 56
请你统计11月份日人均回答量(回答问题数量/答题人数),按回答日期排序,结果保留两位小数,以上例子的输出结果如下:
answer_date per_num
2021-11-01 1.40
2021-11-02
2.00
2021-11-03
1.00
2021-11-04
2.00
2021-11-05
1.25

示例1

输入:

drop table if exists answer_tb;
CREATE TABLE answer_tb(
answer_date date NOT NULL, 
author_id int(10) NOT NULL,
issue_id char(10) NOT NULL,
char_len int(10) NOT NULL);
INSERT INTO answer_tb VALUES('2021-11-1', 101, 'E001' ,150);
INSERT INTO answer_tb VALUES('2021-11-1', 101, 'E002', 200);
INSERT INTO answer_tb VALUES('2021-11-1',102, 'C003' ,50);
INSERT INTO answer_tb VALUES('2021-11-1' ,103, 'P001', 35);
INSERT INTO answer_tb VALUES('2021-11-1', 104, 'C003', 120);
INSERT INTO answer_tb VALUES('2021-11-1' ,105, 'P001', 125);
INSERT INTO answer_tb VALUES('2021-11-1' , 102, 'P002', 105);
INSERT INTO answer_tb VALUES('2021-11-2',  101, 'P001' ,201);
INSERT INTO answer_tb VALUES('2021-11-2',  110, 'C002', 200);
INSERT INTO answer_tb VALUES('2021-11-2',  110, 'C001', 225);
INSERT INTO answer_tb VALUES('2021-11-2' , 110, 'C002', 220);
INSERT INTO answer_tb VALUES('2021-11-3', 101, 'C002', 180);
INSERT INTO answer_tb VALUES('2021-11-4' ,109, 'E003', 130);
INSERT INTO answer_tb VALUES('2021-11-4', 109, 'E001',123);
INSERT INTO answer_tb VALUES('2021-11-5', 108, 'C001',160);
INSERT INTO answer_tb VALUES('2021-11-5', 108, 'C002', 120);
INSERT INTO answer_tb VALUES('2021-11-5', 110, 'P001', 180);
INSERT INTO answer_tb VALUES('2021-11-5' , 106, 'P002' , 45);
INSERT INTO answer_tb VALUES('2021-11-5' , 107, 'E003', 56);

输出:

2021-11-01|1.40
2021-11-02|2.00
2021-11-03|1.00
2021-11-04|2.00
2021-11-05|1.25

原站题解

上次编辑到这里,代码来自缓存 点击恢复默认模板

Mysql 解法, 执行用时: 36ms, 内存消耗: 6380KB, 提交时间: 2022-01-01

select distinct answer_date,round(COUNT(issue_id)/count(DISTINCT author_id),2)
from answer_tb a
where month(answer_date)=11
group by answer_date
order by answer_date

Mysql 解法, 执行用时: 37ms, 内存消耗: 6380KB, 提交时间: 2021-12-21

select answer_date,round(COUNT(issue_id)/count(DISTINCT author_id),2)
from answer_tb a
where month(answer_date)=11
group by answer_date

Mysql 解法, 执行用时: 37ms, 内存消耗: 6388KB, 提交时间: 2021-12-15

select answer_date,round(count(issue_id) /count( distinct author_id),2)  as per_num
FROM answer_tb
group by answer_date

Mysql 解法, 执行用时: 37ms, 内存消耗: 6388KB, 提交时间: 2021-12-08

SELECT answer_date,round(count(issue_id)/count(distinct author_id),2)
from answer_tb
where MONTH(answer_date)=11
group by answer_date
order by answer_date

Mysql 解法, 执行用时: 37ms, 内存消耗: 6388KB, 提交时间: 2021-12-02

select answer_date,round((count(issue_id)/count(distinct(author_id))),2) as per_num from answer_tb group by answer_date