SQL194 某乎问答最大连续回答问题天数大于等于3天的用户及其对应等级

题目描述

现有某乎问答创作者信息表author_tb如下(其中author_id表示创作者编号、author_level表示创作者级别，共1-6六个级别、sex表示创作者性别)：

author_id	author_level	sex
101	6	m
102	1	f
103	1	m
104	3	m
105	4	f
106	2	f
107	2	m
108	5	f
109	6	f
110	5	m

创作者回答情况表answer_tb如下（其中answer_date表示回答日期、author_id指创作者编号、issue_id指回答问题编号、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

请你统计最大连续回答问题的天数大于等于3天的用户及其等级（若有多条符合条件的数据，按author_id升序排序），以上例子的输出结果如下：

author_id	author_level	days_cnt
101	6	3

SQL Schema

drop table if exists author_tb;
CREATE TABLE author_tb(
author_id int(10) NOT NULL, 
author_level int(10) NOT NULL,
sex char(10) NOT NULL);
INSERT INTO author_tb VALUES(101 , 6, 'm');
INSERT INTO author_tb VALUES(102 , 1, 'f');
INSERT INTO author_tb VALUES(103 , 1, 'm');
INSERT INTO author_tb VALUES(104 , 3, 'm');
INSERT INTO author_tb VALUES(105 , 4, 'f');
INSERT INTO author_tb VALUES(106 , 2, 'f');
INSERT INTO author_tb VALUES(107 , 2, 'm');
INSERT INTO author_tb VALUES(108 , 5, 'f');
INSERT INTO author_tb VALUES(109 , 6, 'f');
INSERT INTO author_tb VALUES(110 , 5, 'm');

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);

答案

SELECT author_id, author_level, days_cnt
FROM (SELECT author_id, COUNT(*) AS `days_cnt`
      FROM (SELECT a1.author_id,
                   DATE_SUB(a1.answer_date, INTERVAL
                            ROW_NUMBER() OVER (PARTITION BY a1.author_id ORDER BY a1.answer_date)
                            DAY) `diff`
            FROM (SELECT DISTINCT author_id, answer_date FROM answer_tb) a1) b
      GROUP BY author_id, diff
      HAVING days_cnt >= 3) c
         INNER JOIN author_tb USING (author_id)
ORDER BY author_id;