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SQL(面试实战07)
2022-08-02 10:50:00 【m0_67401499】
1. 某乎问答11月份日人均回答量
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
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);
题解
select
answer_date,
round(count(issue_id)/count(distinct author_id),2) per_num
from answer_tb
group by answer_date
2. 某乎问答高质量的回答中用户属于各级别的数量
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表示创作日期、author_id指创作者编号、issue_id指问题编号、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
回答字数大于等于100字的认为是高质量回答，请你统计某乎问答高质量的回答中用户属于1-2级、3-4级、5-6级的数量分别是多少，按数量降序排列，以上例子的输出结果如下：
level_cut
num
5-6级
12
3-4级
2
1-2级
1
示例1
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);
题解
select level_cut,count(*) num
from
(
select
author_id,author_level,issue_id,
case
when author_level <3 then "1-2级"
when author_level <5 then "3-4级"
else "5-6级"
end
level_cut
from
(
select author_id,author_level,issue_id
from author_tb
join answer_tb
using(author_id)
where char_len>=100
)t
)t1
group by level_cut
order by num desc
3. 某乎问答单日回答问题数大于等于3个的所有用户
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11月份单日回答问题数大于等于3个的所有用户信息（author_date表示回答日期、author_id表示创作者id，answer_cnt表示回答问题个数），以上例子的输出结果如下：
answer_date
author_id
answer_cnt
2021-11-02
110
3
注：若有多条数据符合条件，按answer_date、author_id升序排序。
示例1
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);
题解
select answer_date,author_id,count(*) answer_cnt
from answer_tb
group by answer_date,author_id
having answer_cnt>=3
order by answer_date,author_id
4. 某乎问答回答过教育类问题的用户里有多少用户回答过职场类问题
现有某乎问答题目信息表issue_tb如下（其中issue_id代表问题编号，issue_type表示问题类型）：
issue_id
issue_type
E001
Education
E002
Education
E003
Education
C001
Career
C002
Career
C003
Career
C004
Career
P001
Psychology
P002
Psychology
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
请你统计回答过教育类问题的用户里有多少用户回答过职场类问题，以上例子的输出结果如下：
num
1
示例1
drop table if exists issue_tb;
CREATE TABLE issue_tb(
issue_id char(10) NOT NULL,
issue_type char(10) NOT NULL);
INSERT INTO issue_tb VALUES('E001' ,'Education');
INSERT INTO issue_tb VALUES('E002' ,'Education');
INSERT INTO issue_tb VALUES('E003' ,'Education');
INSERT INTO issue_tb VALUES('C001', 'Career');
INSERT INTO issue_tb VALUES('C002', 'Career');
INSERT INTO issue_tb VALUES('C003', 'Career');
INSERT INTO issue_tb VALUES('C004', 'Career');
INSERT INTO issue_tb VALUES('P001' ,'Psychology');
INSERT INTO issue_tb VALUES('P002' ,'Psychology');
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);
题解
select count(distinct author_id) num
from issue_tb t1
join answer_tb t2
on t1.issue_id=t2.issue_id
where issue_type = 'Career'
and author_id in (
select author_id
from issue_tb a
join answer_tb b
on a.issue_id=b.issue_id
where issue_type = 'Education'
)
5. 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
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表示创作日期、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
示例1
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);
输出
101|6|3
题解
方式一：
-- 最大连续回答问题的天数大于等于3天的用户及其等级
select author_id,author_level,max(rk) days_cnt
from
(
select
author_id,author_level,dense_rank()over(partition by author_id,base_date order by answer_date) rk
from
(
select
author_id,author_level,answer_date,date_sub(answer_date, interval rk day) base_date
from
(
select
author_id,
author_level,
answer_date,
dense_rank()over(partition by author_id order by answer_date) rk
from author_tb
join answer_tb
using(author_id)
)t_rank
)t_row_rank
)t
where rk>=3
group by author_id,author_level
order by author_id
方式二：
select
author_id,author_level,count(distinct answer_date)days_count
from
(
select
author_id,
author_level,
answer_date,
dense_rank()over(partition by author_id order by answer_date) rk
from author_tb
join answer_tb
using(author_id)
)t_rank
group by author_id,author_level,DATE_ADD(answer_date, interval -rk day)
having days_count >=3
order by author_id
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