经典面试题及答案,MySQL去除重复数据

SQL2010不能附加数据库,提示“不能展现央求的对话框”(nColIndex实际值是-1)图像和文字消除方法

前日越过三个急需对表实行去重的主题材料,数据量差不离千万左右,第黄金时代选取正是按Oracle的思绪上:

SQL数据库面试题以至答案

Student(Sno,Sname,Sage,Ssex)
学子表       Sno:学号;Sname:学子姓名;Sage:学子年龄;Ssex:学子性别
Course(Cno,Cname,Tno)
课程表                    Cno,课程编号;Cname:课程名字;Tno:教师编号
SC(Sno,Cno,score)
成绩表                             Sno:学号;Cno,课程编号;score:成绩
Teacher(Tno,Tname) 教授表                        Tno:教师编号;
Tname:教授名字

 

创办数据表:

create table Student(

Sno varchar(10) not null primary key,

Sname varchar(20) null,

Sage int null,

Ssex varchar(2) null

)

create table Course(

Cno varchar(10) not null primary key,

Cname varchar(20) null,

Tno varchar(20) null

)

create table SC(

Sno varchar(10) not null primary key,

Cno varchar(10) not null primary key,

score int null

)

create table Teacher(

Tno varchar(20) primary key not null,

Tname varchar(20) null

)

问题:
1、查询“001”课程比“002”课程战表高的具有学员的学号;
  select a.Sno from (select Sno,score from SC where Cno=’001′) a,(select
Sno,score
  from SC where Cno=’002′) b
  where a.score>b.score and a.Sno=b.Sno;
2、查询平均战表抢先60分的同桌的学号和平均战表;
    select Sno,avg(score)
    from sc
    group by Sno having avg(score) >60;
3、查询全数同学的学号、姓名、选课数、总成绩;
  select Student.Sno,Student.Sname,count(SC.Cno),sum(score)
  from Student left Outer join SC on Student.Sno=SC.Sno
  group by Student.Sno,Sname
4、查询姓“李”的教师职员和工人的个数;
  select count(distinct(Tname))
  from Teacher
  where Tname like ‘李%’;
5、查询没学过“叶平”老师课的同学的学号、姓名;
    select Student.Sno,Student.Sname
    from Student 
    where Sno not in (select distinct( SC.Sno) from SC,Course,Teacher
where SC.Cno=Course.Cno and Teacher.Tno=Course.Tno  and
Teacher.Tname=’叶平’);
6、查询学过“001”而且也学过数码“002”课程的同桌的学号、姓名;
  select Student.Sno,Student.Sname from Student,SC where
Student.Sno=SC.Sno and SC.Cno=’001’and exists( Select * from SC as
SC_2 where SC_2.Sno=SC.Sno and SC_2.Cno=’002′);
7、查询学过“叶平”老师所教的全部课的同学的学号、姓名;
  select Sno,Sname
  from Student
  where Sno in (select Sno from SC ,Course ,Teacher where
SC.Cno=Course.Cno and Teacher.Tno=Course.Tno and Teacher.Tname=’叶平’
group by Sno having count(SC.Cno)=(select count(Cno) from
Course,Teacher  where Teacher.Tno=Course.Tno and Tname=’叶平’));
8、查询课程编号“002”的成就比课程编号“001”课程低的富有同学的学号、姓名;
  Select Sno,Sname from (select Student.Sno,Student.Sname,score ,(select
score from SC SC_2 where SC_2.Sno=Student.Sno and SC_2.Cno=’002′)
score2
  from Student,SC where Student.Sno=SC.Sno and Cno=’001′) S_2 where
score2 <score;
9、查询全数科目战表小于60分的同班的学号、姓名;
  select Sno,Sname
  from Student
  where Sno not in (select Student.Sno from Student,SC where
S.Sno=SC.Sno and score>60);
10、查询未有学全全数课的同校的学号、姓名;
    select Student.Sno,Student.Sname
    from Student,SC
    where Student.Sno=SC.Sno group by  Student.Sno,Student.Sname having
count(Cno) <(select count(Cno) from Course);

11、查询至罕见一门课与学号为“1001”的同班所学相近的同班的学号和人名;
    select Sno,Sname from Student,SC where Student.Sno=SC.Sno and Cno in
select Cno from SC where Sno=’1001′;
12、查询最少学过学号为“001”同学全数一门课的别的同学学号和人名;
    select distinct SC.Sno,Sname
    from Student,SC
    where Student.Sno=SC.Sno and Cno in (select Cno from SC where
Sno=’001′);
13、把“SC”表中“叶平”老师教的课的成就都转移为此课程的平均成绩;
    update SC set score=(select avg(SC_2.score)
    from SC SC_2
    where SC_2.Cno=SC.Cno ) from Course,Teacher where Course.Cno=SC.Cno
and Course.Tno=Teacher.Tno and Teacher.Tname=’叶平’);
14、查询和“1002”号的同窗学习的科目完全相似的别的同学学号和人名;
    select Sno from SC where Cno in (select Cno from SC where
Sno=’1002′)
    group by Sno having count(*)=(select count(*) from SC where
Sno=’1002′);
15、删除学习“叶平”老师课的SC表记录;
    Delect SC
    from course ,Teacher 
    where Course.Cno=SC.Cno and Course.Tno= Teacher.Tno and
Tname=’叶平’;
16、向SC表中插入一些记录,那么些记录供给切合以下准绳:未有上过编号“003”课程的校友学号、2、
    号课的平分战表;
    Insert SC select Sno,’002′,(Select avg(score)
    from SC where Cno=’002′) from Student where Sno not in (Select Sno
from SC where Cno=’002′);
17、按平均成绩从高到低展现全部学员的“数据库”、“公司管理”、“克罗地亚共和国(Republic of Croatia)语”三门的教程成绩,按如下方式呈现:
学子ID,,数据库,集团管理,立陶宛共和国(Republic of Lithuania)语,有效课程数,有效平均分
    SELECT Sno as 学生ID
        ,(SELECT score FROM SC WHERE SC.Sno=t.Sno AND Cno=’004′) AS
数据库
        ,(SELECT score FROM SC WHERE SC.Sno=t.Sno AND Cno=’001′) AS
企业管理
        ,(SELECT score FROM SC WHERE SC.Sno=t.Sno AND Cno=’006′) AS
英语
        ,COUNT(*) AS 有效课程数, AVG(t.score) AS 平均成绩
    FROM SC AS t
    GROUP BY Sno
    ORDER BY avg(t.score) 
18、查询各科战表最高和压低的分:以如下方式显得:课程ID,最高分,最低分
    SELECT L.Cno As 课程ID,L.score AS 最高分,R.score AS 最低分
    FROM SC L ,SC AS R
    WHERE L.Cno = R.Cno and
        L.score = (SELECT MAX(IL.score)
                      FROM SC AS IL,Student AS IM
                      WHERE L.Cno = IL.Cno and IM.Sno=IL.Sno
                      GROUP BY IL.Cno)
        AND
        R.Score = (SELECT MIN(IR.score)
                      FROM SC AS IR
                      WHERE R.Cno = IR.Cno
                  GROUP BY IR.Cno
                    );
19、按各科平均成绩从低到高和及格率的比例从高到低依次
    SELECT t.Cno AS 课程号,max(course.Cname)AS
课程名,isnull(AVG(score),0) AS 平均成绩
        ,100 * SUM(CASE WHEN  isnull(score,0)>=60 THEN 1 ELSE 0
END)/COUNT(*) AS 及格百分数
    FROM SC T,Course
    where t.Cno=course.Cno
    GROUP BY t.Cno
    ORDER BY 100 * SUM(CASE WHEN  isnull(score,0)>=60 THEN 1 ELSE 0
END)/COUNT(*) DESC
20、查询如下课程平均战表和及格率的比重(用”1行”呈现):
集团管理(001),马克思(002),OO&UML (003),数据库(004)
    SELECT SUM(CASE WHEN Cno =’001′ THEN score ELSE 0 END)/SUM(CASE Cno
WHEN ‘001’ THEN 1 ELSE 0 END) AS 公司处理平均分
        ,100 * SUM(CASE WHEN Cno = ‘001’ AND score >= 60 THEN 1 ELSE
0 END)/SUM(CASE WHEN Cno = ‘001’ THEN 1 ELSE 0 END) AS
企管及格百分数
        ,SUM(CASE WHEN Cno = ‘002’ THEN score ELSE 0 END)/SUM(CASE Cno
WHEN ‘002’ THEN 1 ELSE 0 END) AS 马克思平均分
        ,100 * SUM(CASE WHEN Cno = ‘002’ AND score >= 60 THEN 1 ELSE
0 END)/SUM(CASE WHEN Cno = ‘002’ THEN 1 ELSE 0 END) AS
马克思及格百分数
        ,SUM(CASE WHEN Cno = ‘003’ THEN score ELSE 0 END)/SUM(CASE Cno
WHEN ‘003’ THEN 1 ELSE 0 END) AS UML平均分
        ,100 * SUM(CASE WHEN Cno = ‘003’ AND score >= 60 THEN 1 ELSE
0 END)/SUM(CASE WHEN Cno = ‘003’ THEN 1 ELSE 0 END) AS UML及格百分数
        ,SUM(CASE WHEN Cno = ‘004’ THEN score ELSE 0 END)/SUM(CASE Cno
WHEN ‘004’ THEN 1 ELSE 0 END) AS 数据库平均分
        ,100 * SUM(CASE WHEN Cno = ‘004’ AND score >= 60 THEN 1 ELSE
0 END)/SUM(CASE WHEN Cno = ‘004’ THEN 1 ELSE 0 END) AS
数据库及格百分数
  FROM SC
21、查询分裂老师所教区别学科平均分从高到低彰显
  SELECT max(Z.Tno) AS 教授ID,MAX(Z.Tname) AS 教师姓名,C.Cno AS
课程ID,MAX(C.Cname) AS 课程名称,AVG(Score) AS 平均成绩
    FROM SC AS T,Course AS C ,Teacher AS Z
    where T.Cno=C.Cno and C.Tno=Z.Tno
  GROUP BY C.Cno
  ORDER BY AVG(Score) DESC
22、查询如下课程战表第 3 名到第 6
名的上学的小孩子战绩单:公司处理(001),马克思(002),UML
(003),数据库(004)
    [学生ID],[学子姓名],集团处理,Marx,UML,数据库,平均成绩
    SELECT  DISTINCT top 3
      SC.Sno As 学子学号,
        Student.Sname AS 学子姓名 ,
      T1.score AS 公司管理,
      T2.score AS 马克思,
      T3.score AS UML,
      T4.score AS 数据库,
      ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) +
ISNULL(T4.score,0) as 总分
      FROM Student,SC  LEFT JOIN SC AS T1
                      ON SC.Sno = T1.Sno AND T1.Cno = ‘001’
            LEFT JOIN SC AS T2
                      ON SC.Sno = T2.Sno AND T2.Cno = ‘002’
            LEFT JOIN SC AS T3
                      ON SC.Sno = T3.Sno AND T3.Cno = ‘003’
            LEFT JOIN SC AS T4
                      ON SC.Sno = T4.Sno AND T4.Cno = ‘004’
      WHERE student.Sno=SC.Sno and
      ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) +
ISNULL(T4.score,0)
      NOT IN
      (SELECT
            DISTINCT
            TOP 15 WITH TIES
            ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0)

  • ISNULL(T4.score,0)
          FROM sc
                LEFT JOIN sc AS T1
                          ON sc.Sno = T1.Sno AND T1.Cno = ‘k1’
                LEFT JOIN sc AS T2
                          ON sc.Sno = T2.Sno AND T2.Cno = ‘k2’
                LEFT JOIN sc AS T3
                          ON sc.Sno = T3.Sno AND T3.Cno = ‘k3’
                LEFT JOIN sc AS T4
                          ON sc.Sno = T4.Sno AND T4.Cno = ‘k4’
          ORDER BY ISNULL(T1.score,0) + ISNULL(T2.score,0) +
    ISNULL(T3.score,0) + ISNULL(T4.score,0) DESC);

23、计算列印各科战绩,各分数段人数:课程ID,课程名称,[100-85],[85-70],[70-60],[
<60]
    SELECT SC.Cno as 课程ID, Cname as 课程名称
        ,SUM(CASE WHEN score BETWEEN 85 AND 100 THEN 1 ELSE 0 END) AS
[100 – 85]
        ,SUM(CASE WHEN score BETWEEN 70 AND 85 THEN 1 ELSE 0 END) AS
[85 – 70]
        ,SUM(CASE WHEN score BETWEEN 60 AND 70 THEN 1 ELSE 0 END) AS
[70 – 60]
        ,SUM(CASE WHEN score < 60 THEN 1 ELSE 0 END) AS [60 -]
    FROM SC,Course
    where SC.Cno=Course.Cno
    GROUP BY SC.Cno,Cname;

24、查询学子平均成绩及其排名
      SELECT 1+(SELECT COUNT( distinct 平均战绩)
              FROM (SELECT Sno,AVG(score) AS 平均成绩
                      FROM SC
                  GROUP BY Sno
                  ) AS T1
            WHERE 平均成绩 > T2.平分战绩) as 排行,
      Sno as 学生学号,平均战表
    FROM (SELECT Sno,AVG(score) 平均战表
            FROM SC
        GROUP BY Sno
        ) AS T2
    O帕杰罗DE福睿斯 BY 平均成绩 desc;
 
25、查询各科成绩前三名的记录:(不思虑成绩并列景况)
      SELECT t1.Sno as 学生ID,t1.Cno as 课程ID,Score as 分数
      FROM SC t1
      WHERE score IN (SELECT TOP 3 score
              FROM SC
              WHERE t1.Cno= Cno
            ORDER BY score DESC
              )
      ORDER BY t1.Cno;
26、查询每门课程被选修的学子数
  select Cno,count(Sno) from sc group by Cno;
27、查询出只选修了一门学科的整套学子的学号和人名
  select SC.Sno,Student.Sname,count(Cno) AS 选课数
  from SC ,Student
  where SC.Sno=Student.Sno group by SC.Sno ,Student.Sname having
count(Cno)=1;
28、查询男子、女人人数
    Select count(Ssex) as 男士人数 from Student group by Ssex having
Ssex=’男’;
    Select count(Ssex) as 女孩子人数 from Student group by Ssex having
Ssex=’女’;
29、查询姓“张”的学子名单
    SELECT Sname FROM Student WHERE Sname like ‘张%’;
30、查询同名同性别学生名单,并总计同有名的人数
  select Sname,count(*) from Student group by Sname
having  count(*)>1;;
31、1985年诞生的学子名单(注:Student表中Sage列的档期的顺序是datetime)
    select Sname,  CONVERT(char (11),DATEPART(year,Sage)) as age
    from student
    where  CONVERT(char(11),DATEPART(year,Sage))=’1981′;
32、查询每门学科的平分战绩,结果按平均成绩升序排列,平均战绩同样时,按学科号降序排列
    Select Cno,Avg(score) from SC group by Cno order by Avg(score),Cno
DESC ;
33、查询平均成绩超乎85的富有学员的学号、姓名和平均战表
    select Sname,SC.Sno ,avg(score)
    from Student,SC
    where Student.Sno=SC.Sno group by SC.Sno,Sname
having    avg(score)>85;
34、查询课程名字为“数据库”,且分数低于60的学员姓名和分数
    Select Sname,isnull(score,0)
    from Student,SC,Course
    where SC.Sno=Student.Sno and SC.Cno=Course.Cno
and  Course.Cname=’数据库’and score <60;
35、查询全体学子的选课情状;
    SELECT SC.Sno,SC.Cno,Sname,Cname
    FROM SC,Student,Course
    where SC.Sno=Student.Sno and SC.Cno=Course.Cno ;
36、查询任何一门科目成绩在70分以上的姓名、课程名称和分数;
    SELECT  distinct student.Sno,student.Sname,SC.Cno,SC.score
    FROM student,Sc
    WHERE SC.score>=70 AND SC.Sno=student.Sno;
37、查询不如格的学科,并按学科号从大到小排列
    select Cno from sc where scor e <60 order by Cno ;
38、查询课程编号为003且课程成绩在80分以上的学子的学号和人名;
    select SC.Sno,Student.Sname from SC,Student where SC.Sno=Student.Sno
and Score>80 and Cno=’003′;
39、求选了课程的学习者人数
    select count(*) from sc;
40、查询选修“叶平”老师所授课程的学习者中,成绩最高的学习者姓名及其战绩
    select Student.Sname,score
    from Student,SC,Course C,Teacher
    where Student.Sno=SC.Sno and SC.Cno=C.Cno and C.Tno=Teacher.Tno and
Teacher.Tname=’叶平’ and SC.score=(select max(score)from SC where
Cno=C.Cno );
41、查询各样科目及相应的选修人数
    select count(*) from sc group by Cno;
42、查询分化学科战表相近的上学的儿童的学号、课程号、学子战表
  select distinct  A.Sno,B.score from SC A  ,SC B where A.Score=B.Score
and A.Cno <>B.Cno ;
43、查询每门功战表最佳的前两名
    SELECT t1.Sno as 学生ID,t1.Cno as 课程ID,Score as 分数
      FROM SC t1
      WHERE score IN (SELECT TOP 2 score
              FROM SC
              WHERE t1.Cno= Cno
            ORDER BY score DESC
              )
      ORDER BY t1.Cno;
44、总计每门课程的上学的小孩子选修人数(超过10人的教程才总结)。供给输出课程号和选修人数,查询结果按人头降序排列,查询结果按人口降序排列,若人数相近,按学科号升序排列 
    select  Cno as 课程号,count(*) as 人数
    from  sc 
    group  by  Cno
    order  by  count(*) desc,Cno 
45、检索起码选修两门课程的学员学号
    select  Sno 
    from  sc 
    group  by  Sno
    having  count(*)  >  =  2
46、查询全体上学的小孩子都选修的学科的课程号和课程名
    select  Cno,Cname 
    from  Course 
    where  Cno  in  (select  Cno  from  sc group  by  Cno) 
47、查询没学过“叶平”老师上课的任一门学科的学子姓名
    select Sname from Student where Sno not in (select Sno from
Course,Teacher,SC where Course.Tno=Teacher.Tno and SC.Cno=course.Cno and
Tname=’叶平’);
48、查询两门以上比不上格课程的校友的学号及其平均战绩
    select Sno,avg(isnull(score,0)) from SC where Sno in (select Sno
from SC where score <60 group by Sno having count(*)>2)group by
Sno;
49、检索“004”课程分数小于60,按分数降序排列的同窗学号
    select Sno from SC where Cno=’004’and score <60 order by score
desc;

SQL二〇〇八无法附加数据库,提醒“不可能出示要求的对话框”(nColIndex实际值是-1)图像和文字排除方法 

delete from table t1 where id < (select max(id) from table t2 where t1.c1=t2.c1);  --将c1值相同的记录进行去重,只留下id最大的,写成id>min(id)效果相同。

 

如上相关子查询的SQL在c1上设有索引时间效果与利益率不算低,可是很缺憾MySQL未有这种写法,相通的代表写法在MySQL中功能也低的势如水火,如中间表等手腕。

 

正幸好这个月整理一些shell脚本时管理过mysql导入时出错继续实施的标题,因而测量试验后使用了如下办法:

几眼下,笔者在专门的工作室换了后生可畏台计算机写Code,当自己在叠合数据库的时候,现身如下至极,折腾了半天,依然极其。于是自身就重装SQL,重装之后依旧特别。郁结了,这个时候自家也忘怀去google,傻傻的就把系统重装了(新机)。折腾大约一天以往,系统、SQL、VS二〇〇八、补丁等等全都解决之后,总算“水到渠成了”,这时候,笔者又跑去附加数据库。那回蛋疼了~~还是是败退。于是乎,有了那篇随笔(好记性不比烂笔头):

1.将表数据导出:

 

mysqldump -uroot -p --skip-extended-insert -t DBNAME TABLE>TABLE.sql

然后记一下去重后的记录数:
select count(*) from (select 1 from TABLE group by c1) a;

并发谬误 

2.truncate表,然后创立唯一索引

 

truncate table TABLE;
create unique index IX_c1 on TABLE(c1);

 

3.末段导入数据,须求增加-f选项。

 图片 1

mysql -uroot -p -f DBNAME<TABLE.sql

 

-f的成效是:Continue
even if an SQL error occurs.

以此是由于权力不足导致的,大家亟须在【本地账号】的SQL中,新建三个【域帐号】的SQL顾客,设置达成以后,还亟需手动分配权限。详细的情况如下:

如此那般导入时会报非常多的错误,正是因为唯风流倜傥节制的留存,你只需求最终检查下表的记录数时候与第一步中查到的数码生机勃勃致就足以了。

 

这种去重方式作用比较高,破绽可能是失误时显示器上一批的‘Duplicate
entry’报错会清除别的的报错。

肃清办法

除此以外还足以写存款和储蓄进度来删除重复数据,这种措施对数据库的熏陶相当的小,不必要导出导入数据,存款和储蓄进度写法详见:

 

 图片 2

01、回到你系统的本地账号,然后登入你的SQL,作者那篇文章以报到(.)为例,若是你想登录(.\SQLEXPRESS),方法也同等,不做表明。

 

发表评论

电子邮件地址不会被公开。 必填项已用*标注

标签:
网站地图xml地图