oracle 中有没有类似sql中percent的东东

Question

好像没有，没用过
可以这样
SQL> select * from a;AAA                        BBB       CCC
-------------------- --------- ---------
101                          1
102                          2
103                          3
103                         43
102                          5
aaa                          6         6
AAA                          7         7
1                            1         1
aa
bb                     1         1
aa                           1         1
bb
已选择10行。
SQL> select * from a where rownum <= (select max(rownum)*0.2 from a);AAA                        BBB       CCC
-------------------- --------- ---------
101                          1
102                          2

Answer 1

select * from table where rownum<=(select count(*)*0.1 from table);

Answer 2

对PERCENT_RANK()函数的学习:SQL> select * from t_percent;
COL1       COL2
---------- ----------
A          11
A          12
A          13
A          14
A          15
A          16
A          17
A          18
A          19
A          20
B          11
B          12
B          13
B          14
B          15
B          16
B          17
B          18
B          19
B          20
20 rows selected.不分组获得每行在整个纪录集中位置的百分比(从0到1):
----------------------------------------------
SQL> SELECT x.*,
  2    percent_rank() over (partition by 1 order by rownum) AS pr
  3  FROM t_percent x;
COL1       COL2               PR
---------- ---------- ----------
A          11                  0
A          12         .052631579
A          13         .105263158
A          14         .157894737
A          15         .210526316
A          16         .263157895
A          17         .315789474
A          18         .368421053
A          19         .421052632
A          20         .473684211
B          11         .526315789
B          12         .578947368
B          13         .631578947
B          14         .684210526
B          15         .736842105
B          16         .789473684
B          17         .842105263
B          18         .894736842
B          19         .947368421
B          20                  1
20 rows selected.分组获得每行纪录在相应分组中位置的百分比:
----------------------------------------------
SQL> SELECT x.*,
  2    percent_rank() over (partition by col1 order by rownum) AS pr
  3  FROM t_percent x;
COL1       COL2               PR
---------- ---------- ----------
A          11                  0
A          12         .111111111
A          13         .222222222
A          14         .333333333
A          15         .444444444
A          16         .555555556
A          17         .666666667
A          18         .777777778
A          19         .888888889
A          20                  1
B          11                  0
B          12         .111111111
B          13         .222222222
B          14         .333333333
B          15         .444444444
B          16         .555555556
B          17         .666666667
B          18         .777777778
B          19         .888888889
B          20                  1
20 rows selected.我想这样的结果应该足够回答贴主的问题了.

Answer 3

thanks very much!100 point give you