oracle 语法 over 的使用

SQL> select  deptno, dname,lag(dname,2) over (order   by   dname) lag from   dept;--按dname后退二行显示
 
DEPTNO DNAME          LAG
------ -------------- --------------
    10 ACCOUNTING     
    40 OPERbTIONS     
    20 RESEARCH       ACCOUNTING
    30 SALES          OPERbTIONS

SQL> select   deptno, dname,lead(dname,2)   over   (order   by   dname) lag from   dept;--与上相反
 
DEPTNO DNAME          LAG
------ -------------- --------------
    10 ACCOUNTING     RESEARCH
    40 OPERbTIONS     SALES
    20 RESEARCH       
    30 SALES   
对这个表进行操作
NAME       CLASS           SCORE
---------- ---------- ----------
ff         1                  97
gg         1                  89
ll         1                  96
jj         2                  89
oo         2                  87
ii         1                  98
kk         2                  93
uu         3                  97
rr         3                  95
ee         3                  92
yy         2                  90
mm         4                 100
nn         4                  98
pp         1                  98
SQL> select t.name,t.class,t.score from (select name,class,score ,rank() over ( partition by class order by score desc) m from c_score  )t where m=1;--求各班成绩最高的人
 
NAME       CLASS           SCORE
---------- ---------- ----------
ii         1                  98
pp         1                  98
kk         2                  93
uu         3                  97
mm         4                 100
      --rank()是跳跃排序，有两个第二名时接下来就是第四名
       --dense_rank()l是连续排序，有两个第二名时仍然跟着第三名
SQL> select name,class,score,sum(score) over (partition by class) total from c_score;显示各班总分并显示出SCORE列
 
NAME       CLASS           SCORE      TOTAL
---------- ---------- ---------- ----------
pp         1                  98        478
ii         1                  98        478
ll         1                  96        478
ff         1                  97        478
gg         1                  89        478
oo         2                  87        359
jj         2                  89        359
kk         2                  93        359
yy         2                  90        359
ee         3                  92        284
rr         3                  95        284
uu         3                  97        284
nn         4                  98        198
mm         4                 100        198

SQL> select class,sum(score) from c_score group by class;这样不能显示出score这列
 
CLASS      SUM(SCORE)
---------- ----------
1                 478
3                 284
2                 359
4                 198

SQL> select name,class,score,round(score*100/sum(score) over (partition by class),2)||'%' total from c_score;
 个人分数占班级总分数的百分比
NAME       CLASS           SCORE TOTAL
---------- ---------- ---------- -----------------------------------------
pp         1                  98 20.5%
ii         1                  98 20.5%
ll         1                  96 20.08%
ff         1                  97 20.29%
gg         1                  89 18.62%
oo         2                  87 24.23%
jj         2                  89 24.79%
kk         2                  93 25.91%
yy         2                  90 25.07%
ee         3                  92 32.39%
rr         3                  95 33.45%
uu         3                  97 34.15%
nn         4                  98 49.49%
mm         4                 100 50.51%
 
select name,class,score,sum(score) over (order by score range between 2 preceding and 2 following) sum_range from c_score;对应的sum_range是score-1<=score<=score+2　这段间的和
over（order by salary rows between 2 preceding and 4 following）
每行对应的数据窗口是之前2行，之后4行 
over（order by salary rows between unbounded preceding and unbounded following）;
over（order by salary range between unbounded preceding and unbounded following）;
over(partition by null);
以上三条语句等效






曾经一道面试题当时没有做出来，意思同如下，在这个表中如果class与score相同，就考虑这行数据多余，删除多余行，就随便保留一行。
NAME       CLASS           SCORE
---------- ---------- ----------
ff         1                  97
gg         1                  89
ll         1                  96
jj         2                  89
oo         2                  87
ii         1                  98
kk         2                  93
uu         3                  97
rr         3                  95
ee         3                  92
yy         2                  90
mm         4                 100
nn         4                  98
pp         1                  98
fft        1                  97
ggt        1                  89
oot        2                  87
kkt        2                  93
ffff       1                  97
SQL> delete from c_score t where rowid in(select rowid from (select rowid ,row_number() over (partition by class,score order by class)dup_num from c_score)t where t.dup_num>1);
 
6 rows deleted
 
SQL> select * from c_score;
 
NAME       CLASS           SCORE
---------- ---------- ----------
ff         1                  97
gg         1                  89
ll         1                  96
jj         2                  89
kk         2                  93
uu         3                  97
rr         3                  95
ee         3                  92
yy         2                  90
mm         4                 100
nn         4                  98
pp         1                  98
oot        2                  87