正确答案✅：用group by就可以了哇。

创建测试数据：

ALTER SESSION SET nls_date_format = 'YYYY-MM-DD HH24:MI:SS';
SQL> create table test (id number,date1 date);

Table created.

SQL>
SQL>

SQL> insert into test values(1,sysdate-1/8);

1 row created.

SQL> insert into test values(1,sysdate-1/12);

1 row created.

SQL> insert into test values(1,sysdate-1/14);

1 row created.

SQL> insert into test values(1,sysdate-1/15);

1 row created.

SQL> insert into test values(2,sysdate-1/15);

1 row created.

SQL> insert into test values(2,sysdate-1/13);

1 row created.

SQL> insert into test values(2,sysdate-1/17);

1 row created.

SQL> insert into test values(2,sysdate-1/19);

1 row created.

SQL> insert into test values(3,sysdate-1/19);

1 row created.

SQL> insert into test values(3,sysdate-1/32);

1 row created.

SQL> insert into test values(3,sysdate-1/21);

1 row created.

SQL> insert into test values(3,sysdate-1/8);

1 row created.

SQL>
SQL>
SQL> commit;

Commit complete.

SQL>
SQL>
SQL>
SQL> select * from test;

统计sql：

select id,count(id)
from test
where date1 >= sysdate - 1/8
group by id;

希望能帮助到你！望采纳~