已知表name_age如下： create table name_age(name varchar2(20) not null primary key,age integer); 问题：查询出各年龄段的人数 1.显示如下 10-19 5 20-29 7 30-39 2 . . . . . . 2.显示如下 10-19 20-29 30-39 ... 5 7 2 ...
3.显示如下 11-20 5 21-30 7 31-40 2 . . . . . .
答案:
1,select decode(trunc(age/10),0,'0-10',1,'10-19',....),count(*) num from name_age group by trunc(age/10)
2, select sum(decode(trunc(age/10),0,1,0)) '0-9',select sum(decode(trunc(age/10),1,,1,0)) '10-19', ..... from name_age
3,select decode(trunc((age-1)/10),0,'0-10',1,'11-20',....),count(*) num from name_age group by trunc((age-1)/10)
获取当前时间所在周的起始时间
select trunc(to_date('2009-04-27','yyyy-MM-dd'),'WW') from dual;
