• SQL高级查询


    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、查询课程编号“002”的成绩比课程编号“001”课程低的所有同学的学号、姓名;
     
    Select S#,Sname from (select Student.S#,Student.Sname,score ,(select score from SC SC_2 where SC_2.S#=Student.S# and SC_2.C#='002') score2
     
    from Student,SC where Student.S#=SC.S# and C#='001') S_2 where score2 <score;
    9、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);
    10、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);

    11、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';
    12、查询至少学过学号为“001”同学所有一门课的其他同学学号和姓名;
       
    select distinct SC.S#,Sname
       
    from Student,SC
       
    where Student.S#=SC.S# and C# in (select C# from SC where S#='001');
    13、把“SC”表中“叶平”老师教的课的成绩都更改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.C#=SC.C# ) from Course,Teacher where Course.C#=SC.C# and Course.T#=Teacher.T# and Teacher.Tname='叶平');
    14、查询和“1002”号的同学学习的课程完全相同的其他同学学号和姓名;
       
    select S# from SC where C# in (select C# from SC where S#='1002')
       
    group by S# having count(*)=(select count(*) from SC where S#='1002');
    15、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';
    16、向SC表中插入一些记录,这些记录要求符合以下条件:没有上过编号“003”课程的同学学号、2
        号课的平均成绩;
       
    Insert SC select S#,'002',(Select avg(score)
       
    from SC where C#='002') from Student where S# not in (Select S# from SC where C#='002');
    17、按平均成绩从高到低显示所有学生的“数据库”、“企业管理”、“英语”三门的课程成绩,按如下形式显示: 学生ID,,数据库,企业管理,英语,有效课程数,有效平均分
       
    SELECT S# as 学生ID
            ,(
    SELECT score FROM SC WHERE SC.S#=t.S# AND C#='004') AS 数据库
            ,(
    SELECT score FROM SC WHERE SC.S#=t.S# AND C#='001') AS 企业管理
            ,(
    SELECT score FROM SC WHERE SC.S#=t.S# AND C#='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 S#
       
    ORDER BY avg(t.score) 
    18、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# As 课程ID,L.score AS 最高分,R.score AS 最低分
       
    FROM SC L ,SC AS R
       
    WHERE L.C# = R.C# and
            L.score
    = (SELECT MAX(IL.score)
                         
    FROM SC AS IL,Student AS IM
                         
    WHERE L.C# = IL.C# and IM.S#=IL.S#
                         
    GROUP BY IL.C#)
           
    AND
            R.Score
    = (SELECT MIN(IR.score)
                         
    FROM SC AS IR
                         
    WHERE R.C# = IR.C#
                     
    GROUP BY IR.C#
                        );
    19、按各科平均成绩从低到高和及格率的百分数从高到低顺序
       
    SELECT t.C# 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.C#=course.C#
       
    GROUP BY t.C#
       
    ORDER BY 100 * SUM(CASE WHEN  isnull(score,0)>=60 THEN 1 ELSE 0 END)/COUNT(*) DESC
    20、查询如下课程平均成绩和及格率的百分数(用"1行"显示): 企业管理(001),马克思(002),OO&UML (003),数据库(004
       
    SELECT SUM(CASE WHEN C# ='001' THEN score ELSE 0 END)/SUM(CASE C# WHEN '001' THEN 1 ELSE 0 END) AS 企业管理平均分
            ,
    100 * SUM(CASE WHEN C# = '001' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '001' THEN 1 ELSE 0 END) AS 企业管理及格百分数
            ,
    SUM(CASE WHEN C# = '002' THEN score ELSE 0 END)/SUM(CASE C# WHEN '002' THEN 1 ELSE 0 END) AS 马克思平均分
            ,
    100 * SUM(CASE WHEN C# = '002' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '002' THEN 1 ELSE 0 END) AS 马克思及格百分数
            ,
    SUM(CASE WHEN C# = '003' THEN score ELSE 0 END)/SUM(CASE C# WHEN '003' THEN 1 ELSE 0 END) AS UML平均分
            ,
    100 * SUM(CASE WHEN C# = '003' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '003' THEN 1 ELSE 0 END) AS UML及格百分数
            ,
    SUM(CASE WHEN C# = '004' THEN score ELSE 0 END)/SUM(CASE C# WHEN '004' THEN 1 ELSE 0 END) AS 数据库平均分
            ,
    100 * SUM(CASE WHEN C# = '004' AND score >= 60 THEN 1 ELSE 0 END)/SUM(CASE WHEN C# = '004' THEN 1 ELSE 0 END) AS 数据库及格百分数
     
    FROM SC

    21、查询不同老师所教不同课程平均分从高到低显示
     
    SELECT max(Z.T#) AS 教师ID,MAX(Z.Tname) AS 教师姓名,C.C# 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.C#=C.C# and C.T#=Z.T#
     
    GROUP BY C.C#
     
    ORDER BY AVG(Score) DESC
    22、查询如下课程成绩第 3 名到第 6 名的学生成绩单:企业管理(001),马克思(002),UML (003),数据库(004
       
    [学生ID],[学生姓名],企业管理,马克思,UML,数据库,平均成绩
       
    SELECT  DISTINCT top 3
          SC.S#
    As 学生学号,
            Student.Sname
    AS 学生姓名 ,
          T1.score
    AS 企业管理,
          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.S# = T1.S# AND T1.C# = '001'
               
    LEFT JOIN SC AS T2
                         
    ON SC.S# = T2.S# AND T2.C# = '002'
               
    LEFT JOIN SC AS T3
                         
    ON SC.S# = T3.S# AND T3.C# = '003'
               
    LEFT JOIN SC AS T4
                         
    ON SC.S# = T4.S# AND T4.C# = '004'
         
    WHERE student.S#=SC.S# 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.S# = T1.S# AND T1.C# = 'k1'
               
    LEFT JOIN sc AS T2
                         
    ON sc.S# = T2.S# AND T2.C# = 'k2'
               
    LEFT JOIN sc AS T3
                         
    ON sc.S# = T3.S# AND T3.C# = 'k3'
               
    LEFT JOIN sc AS T4
                         
    ON sc.S# = T4.S# AND T4.C# = 'k4'
         
    ORDER BY ISNULL(T1.score,0) + ISNULL(T2.score,0) + ISNULL(T3.score,0) + ISNULL(T4.score,0) DESC);

    23、统计列印各科成绩,各分数段人数:课程ID,课程名称,[100-85],[85-70],[70-60],[ <60]
       
    SELECT SC.C# 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.C#=Course.C#
       
    GROUP BY SC.C#,Cname;

    24、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) AS 平均成绩
                         
    FROM SC
                     
    GROUP BY S#
                      )
    AS T1
               
    WHERE 平均成绩 > T2.平均成绩) as 名次,
          S#
    as 学生学号,平均成绩
       
    FROM (SELECT S#,AVG(score) 平均成绩
               
    FROM SC
           
    GROUP BY S#
            )
    AS T2
       
    ORDER BY 平均成绩 desc;
     
    25、查询各科成绩前三名的记录:(不考虑成绩并列情况)
         
    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#;
    26、查询每门课程被选修的学生数
     
    select c#,count(S#) from sc group by C#;
    27、查询出只选修了一门课程的全部学生的学号和姓名
     
    select SC.S#,Student.Sname,count(C#) AS 选课数
     
    from SC ,Student
     
    where SC.S#=Student.S# group by SC.S# ,Student.Sname having count(C#)=1;
    28、查询男生、女生人数
       
    Select count(Ssex) as 男生人数 from Student group by Ssex having Ssex='';
       
    Select count(Ssex) as 女生人数 from Student group by Ssex having Ssex=''
    29、查询姓“张”的学生名单
       
    SELECT Sname FROM Student WHERE Sname like '张%';
    30、查询同名同性学生名单,并统计同名人数
     
    select Sname,count(*) from Student group by Sname having  count(*)>1;;
    31、1981年出生的学生名单(注:Student表中Sage列的类型是datetime)
       
    select Sname,  CONVERT(char (11),DATEPART(year,Sage)) as age
       
    from student
       
    where  CONVERT(char(11),DATEPART(year,Sage))='1981';
    32、查询每门课程的平均成绩,结果按平均成绩升序排列,平均成绩相同时,按课程号降序排列
       
    Select C#,Avg(score) from SC group by C# order by Avg(score),C# DESC ;
    33、查询平均成绩大于85的所有学生的学号、姓名和平均成绩
       
    select Sname,SC.S# ,avg(score)
       
    from Student,SC
       
    where Student.S#=SC.S# group by SC.S#,Sname having    avg(score)>85;
    34、查询课程名称为“数据库”,且分数低于60的学生姓名和分数
       
    Select Sname,isnull(score,0)
       
    from Student,SC,Course
       
    where SC.S#=Student.S# and SC.C#=Course.C# and  Course.Cname='数据库'and score <60;
    35、查询所有学生的选课情况;
       
    SELECT SC.S#,SC.C#,Sname,Cname
       
    FROM SC,Student,Course
       
    where SC.S#=Student.S# and SC.C#=Course.C# ;
    36、查询任何一门课程成绩在70分以上的姓名、课程名称和分数;
       
    SELECT  distinct student.S#,student.Sname,SC.C#,SC.score
       
    FROM student,Sc
       
    WHERE SC.score>=70 AND SC.S#=student.S#;
    37、查询不及格的课程,并按课程号从大到小排列
       
    select c# from sc where scor e <60 order by C# ;
    38、查询课程编号为003且课程成绩在80分以上的学生的学号和姓名;
       
    select SC.S#,Student.Sname from SC,Student where SC.S#=Student.S# and Score>80 and C#='003';
    39、求选了课程的学生人数
       
    select count(*) from sc;
    40、查询选修“叶平”老师所授课程的学生中,成绩最高的学生姓名及其成绩
       
    select Student.Sname,score
       
    from Student,SC,Course C,Teacher
       
    where Student.S#=SC.S# and SC.C#=C.C# and C.T#=Teacher.T# and Teacher.Tname='叶平' and SC.score=(select max(score)from SC where C#=C.C# );
    41、查询各个课程及相应的选修人数
       
    select count(*) from sc group by C#;
    42、查询不同课程成绩相同的学生的学号、课程号、学生成绩
     
    select distinct  A.S#,B.score from SC A  ,SC B where A.Score=B.Score and A.C# <>B.C# ;
    43、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#;
    44、统计每门课程的学生选修人数(超过10人的课程才统计)。要求输出课程号和选修人数,查询结果按人数降序排列,查询结果按人数降序排列,若人数相同,按课程号升序排列 
       
    select  C# as 课程号,count(*) as 人数
       
    from  sc 
       
    group  by  C#
       
    order  by  count(*) desc,c# 
    45、检索至少选修两门课程的学生学号
       
    select  S# 
       
    from  sc 
       
    group  by  s#
       
    having  count(*>  =  2
    46、查询全部学生都选修的课程的课程号和课程名
       
    select  C#,Cname 
       
    from  Course 
       
    where  C#  in  (select  c#  from  sc group  by  c#) 
    47、查询没学过“叶平”老师讲授的任一门课程的学生姓名
       
    select Sname from Student where S# not in (select S# from Course,Teacher,SC where Course.T#=Teacher.T# and SC.C#=course.C# and Tname='叶平');
    48、查询两门以上不及格课程的同学的学号及其平均成绩
       
    select S#,avg(isnull(score,0)) from SC where S# in (select S# from SC where score <60 group by S# having count(*)>2)group by S#;
    49、检索“004”课程分数小于60,按分数降序排列的同学学号
       
    select S# from SC where C#='004'and score <60 order by score desc;
    50、删除“002”同学的“001”课程的成绩
    delete from Sc where S#='001'and C#='001';

     

     

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

    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
    ------------------------------------------
    create table #tmp(rq varchar(10),shengfu nchar(1))

    insert into #tmp values('2005-05-09','')
    insert into #tmp values('2005-05-09','')
    insert into #tmp values('2005-05-09','')
    insert into #tmp values('2005-05-09','')
    insert into #tmp values('2005-05-10','')
    insert into #tmp values('2005-05-10','')
    insert into #tmp values('2005-05-10','')

    1)select rq, sum(case when shengfu='' then 1 else 0 end)'',sum(case when shengfu='' then 1 else 0 end)'' from #tmp group by rq
    2) select N.rq,N.勝,M.負 from (
    select rq,勝=count(*) from #tmp where shengfu=''group by rq)N inner join
    (
    select rq,負=count(*) from #tmp where shengfu=''group by rq)M on N.rq=M.rq
    3)select a.col001,a.a1 胜,b.b1 负 from
    (
    select col001,count(col001) a1 from temp1 where col002='' group by col001) a,
    (
    select col001,count(col001) b1 from temp1 where col002='' group by col001) b
    where a.col001=b.col001

    2.请教一个面试中遇到的SQL语句的查询问题
    表中有A B C三列,用SQL语句实现:当A列大于B列时选择A列否则选择B列,当B列大于C列时选择B列否则选择C列。
    ------------------------------------------
    select (case when a>b then a else b end ),
    (
    case when b>c then b esle c end)
    from table_name

    3.面试题:一个日期判断的sql语句?
    请取出tb_send表中日期(SendTime字段)为当天的所有记录?(SendTime字段为datetime型,包含日期与时间)
    ------------------------------------------
    select * from tb where datediff(dd,SendTime,getdate())=0

    4.有一张表,里面有3个字段:语文,数学,英语。其中有3条记录分别表示语文70分,数学80分,英语58分,请用一条sql语句查询出这三条记录并按以下条件显示出来(并写出您的思路): 
       大于或等于80表示优秀,大于或等于60表示及格,小于60分表示不及格。 
           显示格式: 
           语文              数学                英语 
           及格              优秀                不及格   
    ------------------------------------------
    select
    (
    case when 语文>=80 then '优秀'
           
    when 语文>=60 then '及格'
    else '不及格') as 语文,
    (
    case when 数学>=80 then '优秀'
           
    when 数学>=60 then '及格'
    else '不及格') as 数学,
    (
    case when 英语>=80 then '优秀'
           
    when 英语>=60 then '及格'
    else '不及格') as 英语,
    from table

    5.在sqlserver2000中请用sql创建一张用户临时表和系统临时表,里面包含两个字段ID和IDValues,类型都是int型,并解释下两者的区别?
    ------------------------------------------
    用户临时表:create table #xx(ID int, IDValues int)
    系统临时表:
    create table ##xx(ID int, IDValues int)

    区别:
    用户临时表只对创建这个表的用户的Session可见,对其他进程是不可见的.
    当创建它的进程消失时这个临时表就自动删除.

    全局临时表对整个SQL Server实例都可见,但是所有访问它的Session都消失的时候,它也自动删除.

    6.sqlserver2000是一种大型数据库,他的存储容量只受存储介质的限制,请问它是通过什么方式实现这种无限容量机制的。
    ------------------------------------------
    它的所有数据都存储在数据文件中(*.dbf),所以只要文件够大,SQL    Server的存储容量是可以扩大的.

    SQL Server
    2000 数据库有三种类型的文件:

    主要数据文件
    主要数据文件是数据库的起点,指向数据库中文件的其它部分。每个数据库都有一个主要数据文件。主要数据文件的推荐文件扩展名是 .mdf。

    次要数据文件
    次要数据文件包含除主要数据文件外的所有数据文件。有些数据库可能没有次要数据文件,而有些数据库则有多个次要数据文件。次要数据文件的推荐文件扩展名是 .ndf。

    日志文件
    日志文件包含恢复数据库所需的所有日志信息。每个数据库必须至少有一个日志文件,但可以不止一个。日志文件的推荐文件扩展名是 .ldf。

    7.请用一个sql语句得出结果
    从table1,table2中取出如table3所列格式数据,注意提供的数据及结果不准确,只是作为一个格式向大家请教。
    如使用存储过程也可以。

    table1

    月份mon 部门dep 业绩yj
    -------------------------------
    一月份      01      10
    一月份     
    02      10
    一月份     
    03      5
    二月份     
    02      8
    二月份     
    04      9
    三月份     
    03      8

    table2

    部门dep      部门名称dname
    --------------------------------
          01      国内业务一部
         
    02      国内业务二部
         
    03      国内业务三部
         
    04      国际业务部

    table3 (result)

    部门dep 一月份      二月份      三月份
    --------------------------------------
          01      10        null      null
         
    02      10         8        null
         
    03      null       5        8
         
    04      null      null      9

    ------------------------------------------
    1)
    select a.部门名称dname,b.业绩yj as '一月份',c.业绩yj as '二月份',d.业绩yj as '三月份'
    from table1 a,table2 b,table2 c,table2 d
    where a.部门dep = b.部门dep and b.月份mon = '一月份' and
    a.部门dep
    = c.部门dep and c.月份mon = '二月份' and
    a.部门dep
    = d.部门dep and d.月份mon = '三月份' and
    2)
    select a.dep,
    sum(case when b.mon=1 then b.yj else 0 end) as '一月份',
    sum(case when b.mon=2 then b.yj else 0 end) as '二月份',
    sum(case when b.mon=3 then b.yj else 0 end) as '三月份',
    sum(case when b.mon=4 then b.yj else 0 end) as '四月份',
    sum(case when b.mon=5 then b.yj else 0 end) as '五月份',
    sum(case when b.mon=6 then b.yj else 0 end) as '六月份',
    sum(case when b.mon=7 then b.yj else 0 end) as '七月份',
    sum(case when b.mon=8 then b.yj else 0 end) as '八月份',
    sum(case when b.mon=9 then b.yj else 0 end) as '九月份',
    sum(case when b.mon=10 then b.yj else 0 end) as '十月份',
    sum(case when b.mon=11 then b.yj else 0 end) as '十一月份',
    sum(case when b.mon=12 then b.yj else 0 end) as '十二月份',
    from table2 a left join table1 b on a.dep=b.dep

    8.华为一道面试题
    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

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  • 原文地址:https://www.cnblogs.com/ymyglhb/p/1732387.html
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