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牛客928225814号
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电子科技大学
2018
数据分析师
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刷题
牛客928225814号
2021-10-30 09:59
数据分析师
题解 | #插入记录(二)#
insert into exam_record_before_2021 SELECT null,uid,exam_id,start_time,submit_time,score from exam_record where submit_time<"2021-01-01 00:00"
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牛客928225814号
2021-10-30 09:50
数据分析师
题解 | #插入记录(一)#
insert into exam_record(uid,exam_id,start_time,submit_time,score) values (1001,9001,"2021-09-01 22:11:12","2021-09-01 23:01:12",90), (1002,9002,"2021-09-04 07:01:02",null,null)
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牛客928225814号
2021-10-26 11:04
数据分析师
题解 | #实习广场投递简历分析(三)#
select a.job,a.mon1 as first_year_mon,a.cnt1,b.mon2,b.cnt2 from (select job,LEFT(date,7) as mon1,sum(num) as cnt1 from resume_info where year(date)="2025" group by job,mon1)a join (select job,LEFT(date,7) as mon2,sum(num) as cnt2 from resume_info where year(date)="2026" group by job,mon2)b on a.job...
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牛客928225814号
2021-10-26 10:22
数据分析师
题解 | #实习广场投递简历分析(二)#
select job,LEFT(date,7) as month,sum(num) as cnt from resume_info where year(date)="2025" group by job,month order by month desc,cnt desc
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牛客928225814号
2021-10-26 10:14
数据分析师
题解 | #实习广场投递简历分析(一)#
select distinct job,sum(num) over(partition by job) as cnt from resume_info where YEAR(date)=2025 order by cnt desc
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牛客928225814号
2021-10-26 10:10
数据分析师
题解 | #牛客的课程订单分析(七)#
select ifnull(client_name,"GroupBuy") as source ,count(id) from (select a.id,a.is_group_buy,name as client_name from (select *,count(id) over(partition by user_id) as c from order_info oi where date>"2025-10-15" and status="completed" and product_name not in("JS") order by id) a left join client...
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牛客928225814号
2021-10-25 10:59
数据分析师
题解 | #牛客的课程订单分析(四)#
/*select distinct b.user_id,min(b.date),count(b.user_id) from (select a.id, a.user_id,a.product_name,a.status,a.client_id,a.date from (select *,count(user_id) over(partition by user_id) as c from order_info where date>"2025-10-15" and status="completed" and product_name not in ("JS"))a where a.c&...
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牛客928225814号
2021-10-25 10:50
数据分析师
题解 | #牛客的课程订单分析(三)#
/*select oi.id,oi.user_id,oi.product_name,oi.status,oi.client_id,oi.date from order_info oi join (select user_id from (select * from order_info where date>"2025-10-15" and status="completed" and product_name not in ("JS"))a group by user_id having count(user_id)>=2 order by user_id)b on oi....
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牛客928225814号
2021-10-25 10:05
数据分析师
题解 | #牛客的课程订单分析(二)#
select user_id from (select * from order_info where date>"2025-10-15" and status="completed" and product_name not in ("JS"))a group by user_id having count(user_id)>=2 order by user_id
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牛客928225814号
2021-10-25 09:44
数据分析师
题解 | #牛客的课程订单分析(一)#
select * from order_info where date>"2025-10-15" and product_name in ("C++","Java","Python") and status="completed"
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牛客928225814号
2021-10-25 09:39
数据分析师
题解 | #考试分数(四)#
select job,case when a.c%2=1 then round((a.c+1)/2,0) else round(a.c/2,0) end as start, case when a.c%2=1 then round((a.c+1)/2,0) else round(a.c/2+1,0) end as end from (select distinct job,count(job) as c from grade group by job)a order by job
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牛客928225814号
2021-10-25 09:37
数据分析师
题解 | #考试分数(四)#
select job,case when a.c%2=1 then round((a.c+1)/2,0) else round(a.c/2,0) end as start, case when a.c%2=1 then round((a.c+1)/2,0) else round(a.c/2+1,0) end as end from (select distinct job,count(job) as c from grade group by job)a order by job
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牛客928225814号
2021-10-25 08:27
数据分析师
题解 | #考试分数(四)#
select job,case when a.c%2=1 then round((a.c+1)/2,0) else round(a.c/2,0) end as start, case when a.c%2=1 then round((a.c+1)/2,0) else round(a.c/2+1,0) end as end from (select distinct job,count(job) as c from grade group by job)a order by job
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牛客928225814号
2021-10-25 08:06
数据分析师
题解 | #考试分数(三)#
select b.id,b.name,b.score from (select a.id,a.name,a.score,dense_rank() over(partition by a.name order by a.score desc) as rk from (select g.id,l.name,g.score from grade g join language l on g.language_id=l.id) a)b where b.rk<=2
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牛客928225814号
2021-10-25 07:46
数据分析师
题解 | #考试分数(二)#
select g.id,g.job,g.score from grade g join (select DISTINCT job,round(avg(score),3) as avg from grade group by job)a on g.job=a.job and g.score>a.avg
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