• sql语句面试总结


    1.用一条SQL语句 查询出每门课都大于80分的学生姓名  

    name   kecheng   fenshu
    张三     语文       81
    张三     数学       75
    李四     语文       76
    李四     数学       90
    王五     语文       81
    王五     数学       100
    王五     英语       90

    A: select distinct name from table  where  name not in (select distinct name from table where fenshu<=80)


    2.一道SQL语句面试题,关于group by
    表内容:
    2005-05-09 胜
    2005-05-09 胜
    2005-05-09 负
    2005-05-09 负
    2005-05-10 胜
    2005-05-10 负
    2005-05-10 负

    如果要生成下列结果, 该如何写sql语句?

                胜 负
    2005-05-09 2 2
    2005-05-10 1 2
    ------------------------------------------
    使用函数连接:
    SELECT rq,SUM(CASE  WHEN shengfu='胜' THEN 1 ELSE 0 END) 胜,SUM(CASE WHEN shengfu='负' THEN 1 ELSE 0 END) 负 FROM tmp GROUP BY rq;
    外连接:
    SELECT n.rq,n.胜,m.负 FROM (SELECT rq,COUNT(*) 胜 FROM tmp WHERE shengfu='胜' GROUP BY rq) n INNER JOIN (SELECT rq,COUNT(*) 负 FROM tmp WHERE shengfu='负' GROUP BY rq) m ON n.rq=m.rq;
    自连接:
    SELECT a.rq, a.a1 胜,b.b1 负 FROM
    (SELECT rq,COUNT(rq) a1 FROM tmp WHERE shengfu='胜' GROUP BY rq)a,
    (SELECT rq,COUNT(rq) b1 FROM tmp WHERE shengfu='负' GROUP BY rq)b
     WHERE a.rq=b.rq;
     

    一个表中的Id有多个记录,把所有这个id的记录查出来,并显示共有多少条记录数。
    ------------------------------------------
    select id, Count(*) from tb group by id having count(*)>1
    select * from(select count(ID) as count from table group by ID)T where T.count>1





    SQL资料:和我们在课堂上所学的较大的不同在于:子查询的位置非常灵活,可以出现在from子句后(需要为此查询取别名),
    也可以出现在select的目标列中,请仔细读懂下列例子(有的地方取别名直接就空格后出现别名,有的地方是用as后面跟别名的方式)。
    注:主要是学习基本语句或短语的用法

    Student(S#,Sname,Sage,Ssex) 学生表
    Course(C#,Cname,T#) 课程表
    SC(S#,C#,score) 成绩表
    Teacher(T#,Tname) 教师表

    问题:
    1、查询“001”课程比“002”课程成绩高的所有学生的学号;
    select a.S#
    from (select s#,score from SC where C#=’001′) a,
    (select s#,score from SC where C#=’002′) b
    where a.score>b.score and a.s#=b.s#;

    2、查询平均成绩大于60分的同学的学号和平均成绩;
    select S#,avg(score)
    from sc
    group by S# having avg(score) >60;

    3、查询所有同学的学号、姓名、选课数、总成绩;
    select Student.S#,Student.Sname,count(SC.C#),sum(score)
    from Student left Outer join SC on Student.S#=SC.S#
    group by Student.S#,Sname

    4、查询姓“李”的老师的个数;
    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.S#,Student.Sname
    from Student
    where S# not in (select distinct( SC.S#) from SC,Course,Teacher where SC.C#=Course.C# and Teacher.T#=Course.T# and Teacher.Tname=’叶平’);

    6、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名;
    select Student.S#,Student.Sname
    from Student,SC
    where Student.S#=SC.S# and SC.C#=’001′and exists( Select * from SC as SC_2 where SC_2.S#=SC.S# and SC_2.C#=’002′);
    7、查询学过“叶平”老师所教的所有课的同学的学号、姓名;
    select S#,Sname
    from Student
    where S# in
    (select S# from SC ,Course ,Teacher
    where SC.C#=Course.C# and Teacher.T#=Course.T# and Teacher.Tname=’叶平’ group by S# having count(SC.C#)=(select count(C#)
    from Course,Teacher where Teacher.T#=Course.T# and Tname=’叶平’));

    8、查询所有课程成绩小于60分的同学的学号、姓名;
    select S#,Sname
    from Student
    where S# not in (select Student.S# from Student,SC where S.S#=SC.S# and score>60);

    9、查询没有学全所有课的同学的学号、姓名;
    select Student.S#,Student.Sname
    from Student,SC
    where Student.S#=SC.S#
    group by Student.S#,Student.Sname having count(C#) <(select count(C#) from Course);

    10、查询至少有一门课与学号为“1001”的同学所学相同的同学的学号和姓名;
    select S#,Sname
    from Student,SC
    where Student.S#=SC.S# and C# in (select C# from SC where S#='1001');

    11、删除学习“叶平”老师课的SC表记录;
    Delect SC
    from course ,Teacher
    where Course.C#=SC.C# and Course.T#= Teacher.T# and Tname='叶平';

    12、查询各科成绩最高和最低的分:以如下形式显示:课程ID,最高分,最低分
    SELECT L.C# 课程ID,L.score 最高分,R.score 最低分
    FROM SC L ,SC R
    WHERE L.C# = R.C#
    and
    L.score = (SELECT MAX(IL.score)
    FROM SC IL,Student IM
    WHERE IL.C# = L.C# and IM.S#=IL.S#
    GROUP BY IL.C#)
    and
    R.Score = (SELECT MIN(IR.score)
    FROM SC IR
    WHERE IR.C# = R.C#
    GROUP BY IR.C# );

    13、查询学生平均成绩及其名次
    SELECT 1+(SELECT COUNT( distinct 平均成绩)
    FROM (SELECT S#,AVG(score) 平均成绩
    FROM SC
    GROUP BY S# ) T1
    WHERE 平均成绩 > T2.平均成绩) 名次, S# 学生学号,平均成绩
    FROM (SELECT S#,AVG(score) 平均成绩 FROM SC GROUP BY S# ) T2
    ORDER BY 平均成绩 desc;

    14、查询各科成绩前三名的记录:(考虑成绩并列情况)
    SELECT t1.S# as 学生ID,t1.C# as 课程ID,Score as 分数
    FROM SC t1
    WHERE score IN (SELECT TOP 3 score
    FROM
    (select distinct score from SC order) by score desc)
    ORDER BY t1.C#;

    (2)、查询各科成绩前三名的记录:(bu考虑成绩并列情况)
    SELECT t1.S# as 学生ID,t1.C# as 课程ID,Score as 分数
    FROM SC t1
    WHERE score IN (SELECT TOP 3 score
    FROM SC
    WHERE t1.C#= C#
    ORDER BY score DESC)
    ORDER BY t1.C#;

    15、查询每门功成绩最好的前两名
    SELECT t1.S# as 学生ID,t1.C# as 课程ID,Score as 分数
    FROM SC t1
    WHERE score IN (SELECT TOP 2 score
    FROM SC
    WHERE t1.C#= C#
    ORDER BY score DESC )
    ORDER BY t1.C#;

    补充:
    已经知道原表
    year salary
    ——————
    2000 1000
    2001 2000
    2002 3000
    2003 4000

    解:
    select b.year,sum(a.salary)
    from salary a,salary b
    where a.year<=b.year
    group by b.year
    order by b.year;

    在面试过程中多次碰到一道SQL查询的题目,查询A(ID,Name)表中第31至40条记录,ID作为主键可能是不是连续增长的列,完整的查询语句如下:
    方法一:
    select top 10 *
    from A
    where ID >(select max(ID) from (select top 30 ID from A order by ID ) T) order by ID
    方法二:
    select top 10 *
    from A
    where ID not In (select top 30 ID from A order by ID)
    order by ID

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  • 原文地址:https://www.cnblogs.com/wcyBlog/p/3820652.html
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