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实习广场投递简历分析(三)

[编程题]实习广场投递简历分析(三)
  • 热度指数：89733 时间限制：C/C++ 1秒，其他语言2秒 空间限制：C/C++ 256M，其他语言512M
  • 算法知识视频讲解
在牛客实习广场有很多公司开放职位给同学们投递，同学投递完就会把简历信息存到数据库里。
现在有简历信息表(resume_info)，部分信息简况如下:
id job date num
1 C++ 2025-01-02 53
2 Python 2025-01-02
 23
3 Java 2025-01-02
 12
4 C++ 2025-01-03
 54
5 Python
 2025-01-03
 43
6 Java
 2025-01-03
 41
7 Java
 2025-02-03
 24
8 C++
 2025-02-03
 23
9 Python
 2025-02-03 34
10 Java
 2025-02-04
 42
11
 C++
 2025-02-04
 45
12 Python
 2025-02-04
 59
13
 C++
 2026-01-04
 230
14 Java
 2026-01-04 764
15 Python
 2026-01-04
 644
16 C++
 2026-01-06
 240
17 Java
 2026-01-06
 714
18 Python
 2026-01-06
 624
19 C++
 2026-01-04
 260
20 Java
 2026-02-14 721
21 Python
 2026-02-14
 321
22 C++
 2026-02-14
 134
23 Java
 2026-02-24
 928
24 Python
 2026-02-24
 525
25 C++
 2027-02-06
 231

第1行表示，在2025年1月2号，C++岗位收到了53封简历
......
最后1行表示，在2027年2月6号，C++岗位收到了231封简历

请你写出SQL语句查询在2025年投递简历的每个岗位，每一个月内收到简历的数目，和对应的2026年的同一个月同岗位，收到简历的数目，最后的结果先按first_year_mon月份降序，再按job降序排序显示，以上例子查询结果如下:
job first_year_mon first_year_cnt second_year_mon
 second_year_cnt
Python 2025-02 93 2026-02
 846
Java 2025-02
 66 2026-02
 1649
C++ 2025-02
 68 2026-02
 394
Python
 2025-01
 66 2026-01
 1268
Java
 2025-01
 53 2026-01
 1478
C++
 2025-01
 107 2026-01
 470
解析:
第1行表示Python岗位在2025年2月收到了93份简历，在对应的2026年2月收到了846份简历
......
最后1行表示C++岗位在2025年1月收到了107份简历，在对应的2026年1月收到了470份简历
示例1

输入

drop table if exists resume_info;
CREATE TABLE resume_info (
id int(4) NOT NULL,
job varchar(64) NOT NULL,
date date NOT NULL,
num int(11) NOT NULL,
PRIMARY KEY (id));

INSERT INTO resume_info VALUES
(1,'C++','2025-01-02',53),
(2,'Python','2025-01-02',23),
(3,'Java','2025-01-02',12),
(4,'C++','2025-01-03',54),
(5,'Python','2025-01-03',43),
(6,'Java','2025-01-03',41),
(7,'Java','2025-02-03',24),
(8,'C++','2025-02-03',23),
(9,'Python','2025-02-03',34),
(10,'Java','2025-02-04',42),
(11,'C++','2025-02-04',45),
(12,'Python','2025-02-04',59),
(13,'C++','2026-01-04',230),
(14,'Java','2026-01-04',764),
(15,'Python','2026-01-04',644),
(16,'C++','2026-01-06',240),
(17,'Java','2026-01-06',714),
(18,'Python','2026-01-06',624),
(19,'C++','2026-02-14',260),
(20,'Java','2026-02-14',721),
(21,'Python','2026-02-14',321),
(22,'C++','2026-02-24',134),
(23,'Java','2026-02-24',928),
(24,'Python','2026-02-24',525),
(25,'C++','2027-02-06',231);

输出

Python|2025-02|93|2026-02|846
Java|2025-02|66|2026-02|1649
C++|2025-02|68|2026-02|394
Python|2025-01|66|2026-01|1268
Java|2025-01|53|2026-01|1478
C++|2025-01|107|2026-01|470
算法知识视频讲解
添加回答
Mysql
像初雪一样遇见你头像 像初雪一样遇见你 
with t1 as (
select job,left(date,7) as first_year_mon,sum(num) as first_year_cnt
from resume_info
where year(date)=2025
group by 1,2
),t2 as (
select job,left(date,7) as second_year_mon,sum(num) as second_year_cnt
from resume_info
where year(date)=2026
group by 1,2   
)
select t1.job,first_year_mon,first_year_cnt
,second_year_mon,second_year_cnt
from t1 
left join t2 
on t1.job=t2.job and (left(first_year_mon,4)=left(second_year_mon,4)-1)
and right(first_year_mon,2)=right(second_year_mon,2)
order by 2 desc,1 desc

发表于 2025-10-15 20:03:32 回复(0)
Mysql
牛客620908534号头像 牛客620908534号
select
    job,
    concat('2025-',lpad(month,2,'0')) first_year_mon,
    first_year_cnt,
    concat('2026-',lpad(month,2,'0')) second_year_mon,
    second_year_cnt
from (
    select
        job,month(date) month ,
        sum(case when year(date)=2025 then num end) first_year_cnt,
        sum(case when year(date)=2026 then num end) second_year_cnt
    from
        resume_info
    group by
        job,month
) a
order by
    month desc,job desc
发表于 2025-08-01 16:28:06 回复(0)
Mysql
牛客155605705号头像 牛客155605705号 
with t1 as
(
    select
    job,left(date,7) year_mon,
    sum(num) year_cnt
    from resume_info
    where year(date)=2025
    group by job,left(date,7)
),
t2 as
(
    select
    job,left(date,7) year_mon,
    sum(num) year_cnt
    from resume_info
    where year(date)=2026
    group by job,left(date,7)
)

select
t1.job job,
t1.year_mon first_year_mon,
t1.year_cnt first_year_cnt,
t2.year_mon second_year_mon,
t2.year_cnt second_year_cnt
from t1
join t2
on t1.job=t2.job
and left(t1.year_mon,4)='2025' and left(t2.year_mon,4)='2026'
and right(t1.year_mon,2)=right(t2.year_mon,2)
order by first_year_mon desc, job desc

发表于 2025-07-29 21:12:55 回复(0)
Mysql
牛客247799997号头像 牛客247799997号
创建两个表，第一个表2025年的，第二个26年的，然后联立查询就行。思路要明确。我第一次做错误的一点是，要注意两表联立的时候，单是一个job条件不够，应该更多一个时间条件才能一一对应。
with
    a as(
        select
            job,
            date_format(date,'%Y-%m') first_year_mon,
            sum(num) first_year_cnt
        from
            resume_info
        where
            year(date)='2025'
        group by
            job, date_format(date,'%Y-%m')
        ),
    b as(
        select
            job,
            date_format(date,'%Y-%m') second_year_mon,
            sum(num) second_year_cnt
        from
            resume_info
        where
            year(date)='2026'
        group by
            job, date_format(date,'%Y-%m')
        )
select
    a.job,
    a.first_year_mon,
    a.first_year_cnt,
    b.second_year_mon,
    b.second_year_cnt
from
    a join b on a.job=b.job and right(a.first_year_mon,2)=right(b.second_year_mon,2)
order by
    first_year_mon desc,job desc;

发表于 2025-07-28 21:35:03 回复(0)
Mysql
jacky酱头像 jacky酱
解法一：
select
    a.job,
    a.first_year_mon,
    a.first_year_cnt,
    b.second_year_mon,
    b.second_year_cnt
from (
    select
        job,
        date_format(date,'%Y-%m') as first_year_mon,
        sum(num) as first_year_cnt
    from resume_info r1
    where date between '2025-01-01' and '2025-12-31'
    group by job,first_year_mon
) as a
join (
    select
        job,
        date_format(date,'%Y-%m') as second_year_mon,
        sum(num) as second_year_cnt
    from resume_info r2
    where date between '2026-01-01' and '2026-12-31'
    group by job,second_year_mon
) as b
on a.job = b.job and b.second_year_mon = date_format(date_add(str_to_date(concat(a.first_year_mon,'-01'),'%Y-%m-%d'),interval 12 month),'%Y-%m')
order by first_year_mon desc,job desc;
解法二：
select r1.job, r1.first_year_mon, r1.first_year_cnt, r2.second_year_mon, r2.second_year_cnt
from (
    select
        job, date_format(date, "%Y-%m") as first_year_mon,
        sum(num) as first_year_cnt
    from resume_info
    where year(date) = 2025
    group by job, first_year_mon
) r1
join (
    select
        job, date_format(date, "%Y-%m") as second_year_mon,
        sum(num) as second_year_cnt
    from resume_info
    where year(date) = 2026
    group by job, second_year_mon
) r2 on right(r1.first_year_mon, 2) = right(r2.second_year_mon, 2) and r1.job = r2.job
order by r1.first_year_mon desc, r1.job desc;
发表于 2025-07-22 16:00:48 回复(0)
Mysql
健谈的马来熊等oc头像 健谈的马来熊等oc
这道题为什么大家都是用内连接取交集啊？理论上应该去并集才对吧？因为可能存在某年某个月简历数为0，但是另一年的对应月份有简历
with a0 as(
    select distinct job,date_format(date,'%m') as date_mon
    from resume_info
),
a1 as(
    select job,date_format(date,'%Y-%m') as first_year_mon,sum(num) as first_year_cnt,date_format(date,'%m') as first_mon
    from resume_info
    where date_format(date,'%Y')=2025
    group by job,first_year_mon,first_mon
),
a2 as(
    select job,date_format(date,'%Y-%m') as second_year_mon,sum(num) as second_year_cnt,date_format(date,'%m') as second_mon
    from resume_info
    where date_format(date,'%Y')=2026
    group by job,second_year_mon,second_mon)

    select distinct a0.job,if(first_year_mon is not null,first_year_mon,concat('2025-',a0.date_mon)) as first_year_mon ,    if(first_year_cnt is not null,first_year_cnt,0) as  first_year_cnt,if(second_year_mon is not null,second_year_mon,concat('2026-',a0.date_mon)) as second_year_mon , if(second_year_cnt is not null,second_year_cnt,0) as second_year_cnt
    from a0
    left join a1
    on a0.job=a1.job and a0.date_mon=first_mon
    left join a2
    on a0.job=a2.job and a0.date_mon=second_mon
    order by first_year_mon desc,job desc



发表于 2025-05-05 21:39:54 回复(0)
Mysql
周小果2025头像 周小果2025
select  t1.job  ,first_year_mon,    first_year_cnt, second_year_mon,
second_year_cnt
from
(select  job,substr(date,1,7) first_year_mon,sum(num) first_year_cnt from  resume_info where year(date)='2025'
group by job,substr(date,1,7)   ) t1
  left join(select  job,substr(date,1,7) second_year_mon,sum(num) second_year_cnt from  resume_info where year(date)='2026'
group by job,substr(date,1,7)    )  t2
on substr(t1.first_year_mon,6,2)=substr(t2.second_year_mon,6,2) and t1.job=t2.job
order by first_year_mon desc,job desc
发表于 2025-04-29 16:28:34 回复(0)
Mysql
kukuju头像 kukuju
--分别选出2025 和2026表格 join连接
with t as(
    select job, date,   num,
    left(date,7) as year_mon,
    year(date) as `year`,
    month(date)as mon
    from resume_info
)
,t1 as(
    select job, year_mon as first_year_mon ,sum(num) as first_year_cnt
    from t
    where year=2025
    group by job, first_year_mon
   )
 ,t2 as(  
select job, year_mon as second_year_mon ,sum(num) as second_year_cnt
from t
where year=2026
group by job , second_year_mon
)
select t1.job,first_year_mon,first_year_cnt,second_year_mon,
second_year_cnt
from t1
left join t2
on t1.job=t2.job
where right(first_year_mon,2)=right(second_year_mon,2)
order by first_year_mon desc,job desc
发表于 2025-04-09 18:21:36 回复(0)
Mysql
牛客611160964号头像 牛客611160964号 
with
    k as (
        SELECT
            job,
            DATE_FORMAT (date, '%Y-%m') AS first_year_mon,
            month (date) mon,
            sum(num) first_year_cnt
        FROM
            resume_info
        where
            year (date) = 2025
        group by
            job,
            month (date),
            DATE_FORMAT (date, '%Y-%m')
    ),
    k1 as (
        SELECT
            k.job,
            first_year_mon,
            first_year_cnt,
            DATE_FORMAT (a.date, '%Y-%m') AS second_year_mon,
            month(a.date) mon1,
            sum(a.num) second_year_cnt
        FROM
            resume_info a
        right join k on k.job=a.job and k.mon=month(a.date)
        where
            year (a.date) = 2026
        group by
            k.job,
            k.mon,
            first_year_mon,
            DATE_FORMAT (a.date, '%Y-%m'),
            month(a.date)
    )
    SELECT job,first_year_mon,first_year_cnt, second_year_mon,second_year_cnt from k1
    order by first_year_mon desc,job desc

发表于 2025-04-08 15:05:11 回复(0)
Mysql
海边的微风头像 海边的微风 
with t1 as(
    select job, date_format(date, '%Y-%m') as first_year_mon, sum(num) as first_year_cnt,
    row_number() over(order by date_format(date, '%Y-%m') desc,job desc) as rk
    from resume_info
    where date between '2025-01-01' and '2025-12-31'
    group by job,first_year_mon

),
t2 as(
    select job, date_format(date, '%Y-%m') as second_year_mon, sum(num) as second_year_cnt,
    row_number() over(order by date_format(date, '%Y-%m') desc,job desc) as rk
    from resume_info
    where date between '2026-01-01' and '2026-12-31'
    group by job,second_year_mon
)

select a.job, first_year_mon, first_year_cnt, second_year_mon, second_year_cnt
from t1 as a
inner join t2 as b using(rk)

发表于 2025-03-17 14:31:29 回复(0)
Mysql
内向的华夫饼头像 内向的华夫饼
with t01 as (
    select job as job,
           month(date) as mon,
           year(date)  as year,
           DATE_FORMAT(date, '%Y-%m') as year_mon,
           num as num
        from resume_info
        where year(date) = 2025 or year(date) = 2026
),
t02 as (
    select job      as job,
           year_mon as first_year_mon,
           mon      as mon,
           sum(num) as first_year_cnt
        from t01
        where year = 2025
        group by job,year_mon,mon
),
t03 as (
    select job      as job,
           year_mon as second_year_mon,
           mon      as mon,
           sum(num) as second_year_cnt
        from t01
        where year = 2026
        group by job,year_mon,mon
)
select a.job             as  job,
       a.first_year_mon  as first_year_mon,
       a.first_year_cnt  as first_year_cnt,
       b.second_year_mon as second_year_mon,
       b.second_year_cnt as second_year_cnt
       from t02 a
       left join t03 b
       on a.job = b.job and a.mon = b.mon
       order by first_year_mon desc,job desc;

发表于 2025-03-13 00:20:28 回复(0)
Mysql
牛客169856390号头像 牛客169856390号
用substring函数 
SELECT
	t1.job,
	t1.first_year_mon,
	t1.first_year_cnt,
	t2.second_year_mon,
	t2.second_year_cnt 
FROM
	(
		SELECT
			job,
			SUBSTRING( DATE, 1, 7 ) AS first_year_mon,
			sum( num ) AS first_year_cnt 
		FROM
			resume_info 
		WHERE
			SUBSTRING( DATE, 1, 4 ) = 2025 
		GROUP BY
			job,
			first_year_mon 
		ORDER BY
			first_year_mon DESC,
			job DESC 
	) t1
	LEFT JOIN (
		SELECT
			job,
			SUBSTRING( DATE, 1, 7 ) AS second_year_mon,
			sum( num ) AS second_year_cnt 
		FROM
			resume_info 
		WHERE
			SUBSTRING( DATE, 1, 4 ) = 2026 
		GROUP BY
			job,
			second_year_mon 
		ORDER BY
			second_year_mon DESC,
			job DESC 
	) t2 ON t1.job = t2.job 
where substring(t1.first_year_mon, 6, 7) = substring(t2.second_year_mon, 6, 7)


发表于 2025-02-20 18:32:43 回复(0)
Mysql
coocha头像 coocha
with month_data as
(
select job
,substr(info.date,1,7) as year_mon
,sum(info.num) as mon_cnt
from resume_info info
group by job,substr(info.date,1,7)
)

select d1.job
,d1.year_mon as first_year_mon
,d1.mon_cnt as first_year_cnt
,d2.year_mon as second_year_mon
,d2.mon_cnt as second_year_cnt
from month_data d1
left join month_data d2
on d1.job = d2.job
and substr(d1.year_mon,1,4)+1 = substr(d2.year_mon,1,4)
and substr(d1.year_mon,6,2) = substr(d2.year_mon,6,2)
where substr(d1.year_mon,1,4) = '2025'

order by first_year_mon desc,job desc
发表于 2024-11-28 09:16:00 回复(0)
Mysql
Meraki_Voorpret头像 Meraki_Voorpret
用的窗口函数 但是不知道错哪了 
select a.job, a.first_year_mon, a.first_year_cnt, b.second_year_mon, b.second_year_cnt
from (
    # 2025
    select distinct job, substring(date, 1, 7) as first_year_mon, sum(num) over(partition by job, substring(date, 1, 7) order by substring(date, 1, 7) desc) as first_year_cnt
    from resume_info
    where year(date) = 2025
) as a 
join (
    # 2026
    select distinct job, substring(date, 1, 7) as second_year_mon, sum(num) over(partition by job, substring(date, 1, 7) order by substring(date, 1, 7) desc) as second_year_cnt
    from resume_info
    where year(date) = 2026
) as b on a.job = b.job and date_sub(b.second_year_mon, interval 1 year) = a.first_year_mon
order by a.first_year_mon desc, a.job desc


发表于 2024-11-26 10:32:57 回复(0)
Mysql
牛客621940403号头像 牛客621940403号
--提取2025年資料
with r1 as(
    select 
        job,
        date_format(date , '%Y-%m') AS first_year_mon,
        sum(num) as first_year_cnt
    from resume_info 
    where year(date) = 2025
    group by job, first_year_mon
), r2 as(--提取2026年資料
    select 
        job,
        date_format(date , '%Y-%m') as second_year_mon,
        sum(num) as second_year_cnt
    from resume_info 
    where year(date) = 2026
    group by job, second_year_mon
)
 --將2026年資料加入到2025年資料的旁邊
select 
    r1.job,
    r1.first_year_mon,
    r1.first_year_cnt,
    r2.second_year_mon,
    r2.second_year_cnt
from r1,r2 
    where left(r1.first_year_mon,4) + 1 = left(r2.second_year_mon,4) 
    and right(r1.first_year_mon,2) = right(r2.second_year_mon,2)
    and r1.job = r2.job
order by r1.first_year_mon desc , r1.job desc

发表于 2024-11-19 11:43:29 回复(0)
Mysql
lem328头像 lem328
select fy.job,
    fy.mon,
    fy.cnt,
    sy.mon,
    sy.cnt
from (select a.job,a.mon,
    sum(a.num) as cnt,
    concat(a.job,right(a.mon,2)) as id
    from (select *,
        date_format(date,'%Y-%m') as mon
        from resume_info) as a
    where year(date)=2025
    group by a.job,a.mon) as fy
    join
    (select b.job,b.mon,
    sum(b.num) as cnt,
    concat(b.job,right(b.mon,2)) as id
    from (select *,
        date_format(date,'%Y-%m') as mon
        from resume_info) as b
    where year(date)=2026
    group by b.job,b.mon) as sy
    on fy.id = sy.id
order by fy.mon desc,job desc
发表于 2024-11-12 18:29:38 回复(0)
Mysql
年轻的鲸鱼在春招头像 年轻的鲸鱼在春招
选出2025年及2026年每个月的数据，做两个表之间的连接 
select t1.job,t1.date1 first_year_mon,t1.cnt1 first_year_cnt,
       t2.date1 second_year_mon,t2.cnt1 second_year_cnt
from (
    select job,date1,months,sum(num) cnt1
    from (
            select job,date_format(date,"%Y-%m") date1,
                    num,month(date) as months
            from resume_info
            where year(date)=2025 ) t
    group by job,date1,months
     ) as t1
left join (
    select job,date1,months,sum(num) cnt1
    from (
            select job,date_format(date,"%Y-%m") date1,num,month(date) as months
            from resume_info
            where year(date)=2026 ) t
    group by job,date1,months
     ) as t2
on t1.job = t2.job and t1.months = t2.months
order by t1.date1 desc,t1.job desc
   


发表于 2024-10-22 20:01:21 回复(0)
Mysql
牛客440842856号头像 牛客440842856号
select t1.job,date1 as first_year_mon,first_year_cnt,date2 as second_year_mon,second_year_cnt from
(select job,DATE_FORMAT(date, '%Y-%m') as date1,sum(num) as first_year_cnt from resume_info
where year(date)=2025
group by job,date1) as t1
inner join
(select job,DATE_FORMAT(date, '%Y-%m') as date2,sum(num) as second_year_cnt from resume_info
where year(date)=2026
group by job,date2) as t2
on t1.job=t2.job  
and substring(t2.date2,6,2)=substring(t1.date1,6,2)
order by first_year_mon desc,t1.job desc;
发表于 2024-10-18 17:35:25 回复(0)
Mysql
牛客370141165号头像 牛客370141165号
WITH t as (SELECT job,mon AS first_year_mon,cnt AS first_year_cnt,
       lead(mon,2)OVER(partition by job order by mon) AS second_year_mon,
       lead(cnt,2)OVER(partition by job order by mon) AS second_year_cnt
FROM(
SELECT job, date_format(date,"%Y-%m") AS mon,SUM(num) AS cnt
FROM resume_info
WHERE YEAR(date) in (2025,2026)
group by job,mon
ORDER BY job) AS a)
SELECT * FROM t
where first_year_mon LIKE "2025%"
ORDER BY first_year_mon desc ,job DESC
发表于 2024-10-16 21:26:22 回复(0)
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