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  • 学生各门课程成绩统计SQL语句大全

    学生成绩表(stuscore): 

    姓名:name 课程:subject 分数:score 学号:stuid
    张三 数学 89 1
    张三 语文 80 1
    张三 英语 70 1
    李四 数学 90 2
    李四 语文 70 2
    李四 英语 80 2



    创建表 

    SET ANSI_NULLS ONGOSET QUOTED_IDENTIFIER ONGOSET ANSI_PADDING ONGOCREATE TABLE [dbo].[stuscore](    [name] [varchar](50) COLLATE Chinese_PRC_CI_AS NULL,    [subject] [varchar](50) COLLATE Chinese_PRC_CI_AS NULL,    [score] [int] NULL,    [stuid] [int] NULL) ON [PRIMARY] GOSET ANSI_PADDING OFF



    问题: 

    1.计算每个人的总成绩并排名(要求显示字段:姓名,总成绩) 
    2.计算每个人的总成绩并排名(要求显示字段: 学号,姓名,总成绩) 
    3.计算每个人单科的最高成绩(要求显示字段: 学号,姓名,课程,最高成绩) 
    4.计算每个人的平均成绩(要求显示字段: 学号,姓名,平均成绩) 
    5.列出各门课程成绩最好的学生(要求显示字段: 学号,姓名,科目,成绩) 
    6.列出各门课程成绩最好的两位学生(要求显示字段: 学号,姓名,科目,成绩) 
    7.统计如下: 

    学号 姓名 语文 数学 英语 总分 平均分
                 


    8.列出各门课程的平均成绩(要求显示字段:课程,平均成绩) 
    9.列出数学成绩的排名(要求显示字段:学号,姓名,成绩,排名) 
    10.列出数学成绩在2-3名的学生(要求显示字段:学号,姓名,科目,成绩) 
    11.求出李四的数学成绩的排名 
    12.统计如下: 

    课程 不及格(0-59)个 良(60-80)个 优(81-100)个
           


    13.统计如下:数学:张三(50分),李四(90分),王五(90分),赵六(76分) 

    答案: 


    1.计算每个人的总成绩并排名 

    select name,sum(score) as allscore from stuscore group by name order by allscore


    2.计算每个人的总成绩并排名 

    select distinct t1.name,t1.stuid,t2.allscore from  stuscore t1,(    select stuid,sum(score) as allscore from stuscore group by stuid)t2where t1.stuid=t2.stuidorder by t2.allscore desc


    3. 计算每个人单科的最高成绩 

    select t1.stuid,t1.name,t1.subject,t1.score from stuscore t1,(select stuid,max(score) as maxscore from stuscore group by stuid) t2where t1.stuid=t2.stuid and t1.score=t2.maxscore


    4.计算每个人的平均成绩 

    select distinct t1.stuid,t1.name,t2.avgscore from stuscore t1,(select stuid,avg(score) as avgscore from stuscore group by stuid) t2where t1.stuid=t2.stuid


    5.列出各门课程成绩最好的学生 

    select  t1.stuid,t1.name,t1.subject,t2.maxscore from stuscore t1,(select subject,max(score) as maxscore from stuscore group by subject) t2where t1.subject=t2.subject and t1.score=t2.maxscore


    6.列出各门课程成绩最好的两位学生 

    select distinct t1.* from stuscore t1 where t1.id in (select top 2 stuscore.id from stuscore where subject = t1.subject order by score desc) order by t1.subject


    7.学号     姓名     语文      数学      英语      总分   平均分 

    select stuid as 学号,name as 姓名,sum(case when subject='语文' then score else 0 end) as 语文,sum(case when subject='数学' then score else 0 end) as 数学,sum(case when subject='英语' then score else 0 end) as 英语,sum(score) as 总分,(sum(score)/count(*)) as 平均分from stuscoregroup by stuid,name order by 总分desc


    8.列出各门课程的平均成绩 

    select subject,avg(score) as avgscore from stuscoregroup by subject


    9.列出数学成绩的排名 

    declare @tmp table(pm int,name varchar(50),score int,stuid int)insert into @tmp select null,name,score,stuid from stuscore where subject='数学' order by score descdeclare @id intset @id=0;update @tmp set @id=@id+1,pm=@idselect * from @tmp
    select  DENSE_RANK () OVER(order by score desc) as row,name,subject,score,stuid from stuscore where subject='数学'order by score desc


    10. 列出数学成绩在2-3名的学生 

    select t3.*  from(select top 2 t2.*  from (select top 3 name,subject,score,stuid from stuscore where subject='数学'order by score desc) t2 order by t2.score) t3 order by t3.score desc


    11. 求出李四的数学成绩的排名 

    declare @tmp table(pm int,name varchar(50),score int,stuid int)insert into @tmp select null,name,score,stuid from stuscore where subject='数学' order by score descdeclare @id intset @id=0;update @tmp set @id=@id+1,pm=@idselect * from @tmp where name='李四'


    12. 课程     不及格(-59)    良(-80)    优(-100) 

    select subject, (select count(*) from stuscore where score<60 and subject=t1.subject) as 不及格,(select count(*) from stuscore where score between 60 and 80 and subject=t1.subject) as 良,(select count(*) from stuscore where score >80 and subject=t1.subject) asfrom stuscore t1 group by subject


    13. 数学:张三(50分),李四(90分),王五(90分),赵六(76分) 

    declare @s varchar(1000)set @s=''select @s =@s+','+name+'('+convert(varchar(10),score)+'分)' from stuscore where subject='数学' set @s=stuff(@s,1,1,'')print '数学:'+@s
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  • 原文地址:https://www.cnblogs.com/zhangliang88/p/5430254.html
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