借助rownum中求Oracle表中前三名(三甲:状元榜眼探花)的方法(总计三种方法,以讲述rownum的使用为主)

要求前三名,MySQL中有order by排序,limit限制数量,结果很容易得到,而且limit的执行顺序也在order by之后,写出的sql高效易懂而不易出错。

但在oracle中,由于没有limit子句,人们喜欢求助于rownum伪列,但是,因为rownum身处select子句中,而select子句的执行顺序先于order by,因此只有order by比rownum更深一个层次,这样得到的伪列才有效,否则如果rownum和order by处于同一层次,这时的伪列是无效的!如果忘了这一点,很容易写成错误的sql语句,下面将就此举例说明:

待测试的表结构:

create table tb_student01(
    id number(4,0) not null primary key,
    name nvarchar2(20) not null,
    score number(3,0) not null)

充值语句(注意我故意打乱了插入顺序,就是不想在sql写错时还能因为插入顺序的关系而蒙到正确结果):

insert into tb_student01(id,name,score) values('7','张三',77);
insert into tb_student01(id,name,score) values('8','李四',88);
insert into tb_student01(id,name,score) values('6','Felix',10);
insert into tb_student01(id,name,score) values('10','赵六',15);
insert into tb_student01(id,name,score) values('11','孙七',67);
insert into tb_student01(id,name,score) values('12','钱八',37);
insert into tb_student01(id,name,score) values('1','Andy',67);
insert into tb_student01(id,name,score) values('2','Bill',98);
insert into tb_student01(id,name,score) values('9','王五',100);
insert into tb_student01(id,name,score) values('3','Cindy',25);
insert into tb_student01(id,name,score) values('4','Douglas',64);
insert into tb_student01(id,name,score) values('5','Eliot',99);

稍稍目测一下,100分的王五,99分的Eliot,98分的Bill将是期望的三甲。

首先我们看下面的SQL语句会是什么结果:

select rownum as rn,a.* from tb_student01 a order by a.score desc
SQL> select rownum as rn,a.* from tb_student01 a order by a.score desc;

        RN         ID NAME                                          SCORE
---------- ---------- ---------------------------------------- ----------
         9          9 王五                                            100
        12          5 Eliot                                            99
         8          2 Bill                                             98
         2          8 李四                                             88
         1          7 张三                                             77
         5         11 孙七                                             67
         7          1 Andy                                             67
        11          4 Douglas                                          64
         6         12 钱八                                             37
        10          3 Cindy                                            25
         4         10 赵六                                             15

        RN         ID NAME                                          SCORE
---------- ---------- ---------------------------------------- ----------
         3          6 Felix                                            10

已选择12行。

已用时间:  00: 00: 00.00

从上面我们就能看出王五的rn值不是1而是9,Eliot的rn值不是2而是12,Bill的rn值不是3而是8!

发生这种情况的原因是rownum身处的select子句执行顺序是高于order by的,这导致了排序前rownum就按初始的插入顺序被赋上了值。

再用rn值进行筛选,得到的结果让人瞠目结舌:

select b.* from
(select rownum as rn,a.* from tb_student01 a order by a.score desc) b
where b.rn<4
SQL> select b.* from
  2  (select rownum as rn,a.* from tb_student01 a order by a.score desc) b
  3  where b.rn<4;

        RN         ID NAME                                          SCORE
---------- ---------- ---------------------------------------- ----------
         2          8 李四                                             88
         1          7 张三                                             77
         3          6 Felix                                            10

已用时间:  00: 00: 00.01

三甲中不仅没有90分以上的,连考10分的Felix都被放到了三甲里,学生们如果知道真实原因会写完姓名后就抢先交卷,因为越早被批改,记录就越早出现在数据库中,也就越有可能位列三甲!这是多么荒唐的事情!

为了杜绝这种荒诞,我们必须重写sql,其中要点是先按分数排序再附上伪列。

首先排序:

select a.* from tb_student01 a order by a.score desc
SQL> select a.* from tb_student01 a order by a.score desc;

        ID NAME                                          SCORE
---------- ---------------------------------------- ----------
         9 王五                                            100
         5 Eliot                                            99
         2 Bill                                             98
         8 李四                                             88
         7 张三                                             77
        11 孙七                                             67
         1 Andy                                             67
         4 Douglas                                          64
        12 钱八                                             37
         3 Cindy                                            25
        10 赵六                                             15

        ID NAME                                          SCORE
---------- ---------------------------------------- ----------
         6 Felix                                            10

已选择12行。

已用时间:  00: 00: 00.01

其次,再附上伪列:

SQL> select rownum as rn,b.* from
  2  (select a.* from tb_student01 a order by a.score desc) b ;

        RN         ID NAME                                          SCORE
---------- ---------- ---------------------------------------- ----------
         1          9 王五                                            100
         2          5 Eliot                                            99
         3          2 Bill                                             98
         4          8 李四                                             88
         5          7 张三                                             77
         6         11 孙七                                             67
         7          1 Andy                                             67
         8          4 Douglas                                          64
         9         12 钱八                                             37
        10          3 Cindy                                            25
        11         10 赵六                                             15

        RN         ID NAME                                          SCORE
---------- ---------- ---------------------------------------- ----------
        12          6 Felix                                            10

已选择12行。

已用时间:  00: 00: 00.01

最后,对rn值进行筛选就得到最终结果了:

SQL> select c.* from
  2  ( select rownum as rn,b.* from
  3  (select a.* from tb_student01 a order by a.score desc) b ) c
  4  where c.rn<4;

        RN         ID NAME                                          SCORE
---------- ---------- ---------------------------------------- ----------
         1          9 王五                                            100
         2          5 Eliot                                            99
         3          2 Bill                                             98

已用时间:  00: 00: 00.01

这个结果和预期值是一致的,所以最终查询三甲的SQL是:

select c.* from
( select rownum as rn,b.* from
(select a.* from tb_student01 a order by a.score desc) b ) c
where c.rn<4

当然,不借助rownum也可以得到正确结果,比如说分析函数rank():

SQL:

select b.* from
(select rank() over (order by score desc) as rn,a.* from tb_student01 a ) b
where b.rn<4

执行结果:

SQL> select b.* from
  2  (select rank() over (order by score desc) as rn,a.* from tb_student01 a ) b
  3  where b.rn<4;

        RN         ID NAME                                          SCORE
---------- ---------- ---------------------------------------- ----------
         1          9 王五                                            100
         2          5 Eliot                                            99
         3          2 Bill                                             98

已用时间:  00: 00: 00.00

或是借助row_number函数

SQL:

select b.* from
(select row_number() over (order by score desc) as rn,a.* from tb_student01 a ) b
where b.rn<4

执行结果:

SQL> select b.* from
  2  (select row_number() over (order by score desc) as rn,a.* from tb_student01 a ) b
  3  where b.rn<4;

        RN         ID NAME                                          SCORE
---------- ---------- ---------------------------------------- ----------
         1          9 王五                                            100
         2          5 Eliot                                            99
         3          2 Bill                                             98

已用时间:  00: 00: 00.00

--2020年3月28日--

以上用到的全部SQL:

create table tb_student01(
    id number(4,0) not null primary key,
    name nvarchar2(20) not null,
    score number(3,0) not null)
    

insert into tb_student01(id,name,score) values('7','张三',77);
insert into tb_student01(id,name,score) values('8','李四',88);
insert into tb_student01(id,name,score) values('6','Felix',10);
insert into tb_student01(id,name,score) values('10','赵六',15);
insert into tb_student01(id,name,score) values('11','孙七',67);
insert into tb_student01(id,name,score) values('12','钱八',37);
insert into tb_student01(id,name,score) values('1','Andy',67);
insert into tb_student01(id,name,score) values('2','Bill',98);
insert into tb_student01(id,name,score) values('9','王五',100);
insert into tb_student01(id,name,score) values('3','Cindy',25);
insert into tb_student01(id,name,score) values('4','Douglas',64);
insert into tb_student01(id,name,score) values('5','Eliot',99);

Wrong:
select rownum as rn,a.* from tb_student01 a order by a.score desc

select b.* from
(select rownum as rn,a.* from tb_student01 a order by a.score desc) b
where b.rn<4

Correct:
select a.* from tb_student01 a order by a.score desc

select rownum as rn,b.* from
(select a.* from tb_student01 a order by a.score desc) b 

select c.* from
( select rownum as rn,b.* from
(select a.* from tb_student01 a order by a.score desc) b ) c
where c.rn<4

select rank() over (order by score desc) as rn,a.* from tb_student01 a 

select b.* from
(select rank() over (order by score desc) as rn,a.* from tb_student01 a ) b
where b.rn<4

或是row_number函数
select b.* from
(select row_number() over (order by score desc) as rn,a.* from tb_student01 a ) b
where b.rn<4
原文地址:https://www.cnblogs.com/heyang78/p/12122727.html