oracle行转列练习

 ----------------------第一题---------------------------
 create table STUDENT_SCORE
 (
   name    VARCHAR2(20),
   subject VARCHAR2(20),
   score   NUMBER(4,1)
 )

 insert into student_score (NAME, SUBJECT, SCORE) values ('张三', '语文', 78.0);
 insert into student_score (NAME, SUBJECT, SCORE) values ('张三', '数学', 88.0);
 insert into student_score (NAME, SUBJECT, SCORE) values ('张三', '英语', 98.0);
 insert into student_score (NAME, SUBJECT, SCORE) values ('李四', '语文', 89.0);
 insert into student_score (NAME, SUBJECT, SCORE) values ('李四', '数学', 76.0);
 insert into student_score (NAME, SUBJECT, SCORE) values ('李四', '英语', 90.0);
 insert into student_score (NAME, SUBJECT, SCORE) values ('王五', '语文', 99.0);
 insert into student_score (NAME, SUBJECT, SCORE) values ('王五', '数学', 66.0);
 insert into student_score (NAME, SUBJECT, SCORE) values ('王五', '英语', 91.0);

 --emp
 /*
 显示格式：
        语文              数学                英语
        及格              优秀                不及格
 */
 --方法一
 select  name,
                 (select score from student_score s1 where subject = '语文' and s1.name=s.name) 语文,
                 (select score from student_score s1 where subject = '数学' and s1.name=s.name) 数学,
                 (select score from student_score s1 where subject = '英语' and s1.name=s.name) 英语
   from student_score s group by name

 --方法二 decode
 select s.name,
        sum(decode(subject, '语文',s.score,0)) 语文,
        sum(decode(subject, '数学',s.score,0)) 数学,
        sum(decode(subject, '英语',s.score,0)) 英语
   from student_score s
  group by s.name
 --方法三 case when
 select s.name , sum(case s.subject when '语文' then s.score else 0 end) "语文",
 sum(case s.subject when '数学' then s.score else 0 end) 数学,
 sum(case s.subject when '英语' then s.score else 0 end) 英语
 from student_score s group by s.name
 --方法四
 采用 join表连接的方式

 --判断及格否
 select t.name 名字,
        case
          when t.y between 90 and 100 then
           '优秀'
           when t.y between 60 and 90 then
           '及格'
           when t.y between 0 and 60 then
           '不及格'
        end 语文,
       case
          when t.s between 90 and 100 then
           '优秀'
           when t.s between 60 and 90 then
           '及格'
           when t.s between 0 and 60 then
           '不及格'
        end 数学,
        case
          when t.e between 90 and 100 then
           '优秀'
           when t.e between 60 and 90 then
           '及格'
           when t.e between 0 and 60 then
           '不及格'
        end 英语

   from (

         select s.name,
                  sum(decode(subject, '语文', s.score, 0)) y,
                 sum(decode(subject, '数学', s.score, 0)) s,
                 sum(decode(subject, '英语', s.score, 0)) e
           from student_score s
          group by s.name) t

 -----------------------第二题--------------------------------

 create table test(
    id number(10) primary key,
    type number(10) ,
    t_id number(10),
    value varchar2(5)
 );
 insert into test values(100,1,1,'张三');
 insert into test values(200,2,1,'男');
 insert into test values(300,3,1,'');
 /*
 姓名      性别     年龄
 --------- -------- ----
 张三       男        50
 */

 --方法一
 --
 select listagg(decode(t.type, 1, t.value)) within group(order by value) 姓名,
        listagg(decode(t.type, 2, t.value)) within group(order by value) 性别,
        listagg(decode(t.type, 3, t.value)) within group(order by value) 年龄
   from test t
  group by t.t_id

  --方法二
  select max(decode(t.type, 1, t.value)) 姓名,
        max(decode(t.type, 2, t.value)) 性别,
       max(decode(t.type, 3, t.value))年龄
   from test t group by t.t_id

  --方法三表连接方式
  select * from test

 select *
   from (select value name,t_id from test where type = 1) m1
   join (select value sex,t_id from test where type = 2) m2
     on m1.t_id = m2.t_id

  -------------------------第三题-------------------------

 create table tmp(rq varchar2(10),shengfu varchar2(5))

 insert into tmp values('2005-05-09','胜');
 insert into tmp values('2005-05-09','胜');
 insert into tmp values('2005-05-09','负');
 insert into tmp values('2005-05-09','负');
 insert into tmp values('2005-05-10','胜');
 insert into tmp values('2005-05-10','负');
 insert into tmp values('2005-05-10','负');

 select * from tmp;
           胜 负
 2005-05-09 2 2
 2005-05-10 1 2

  --方法一
  select rq,
         sum(decode(shengfu, '胜', 1, '负', 0)) 胜,
         sum(decode(shengfu, '胜', 0, '负', 1)) 负
    from tmp
   group by rq