insert into table1(id,code) values('11224','KYFAWT');
insert into table1(id,code) values('12215','KYFAHS');
insert into table1(id,code) values('12215','KYFAHS');
insert into table1(id,code) values('12220','KYFAWT');
insert into table1(id,code) values('12501','JHRWSXD');
insert into table1(id,code) values('12220','KYFAHS');
insert into table1(id,code) values('12210','KYFAHS');
insert into table1(id,code) values('12501','KYFAWT');
insert into table1(id,code) values('12501','KYFAWT');
insert into table1(id,code) values('11224','KYFAWT');
insert into table1(id,code) values('12220','KYFAWT');
commit;
select distinct id,count(distinct code) over (partition by id) from table1 order by id;
--结果
1    11224    1
2    12210    1
3    12215    1
4    12220    2
5    12501    2

------解决方案--------------------

SQL code

SELECT ID,
       SUM(CASE CODE WHEN 'KYFAWT' THEN 1 ELSE 0 END) AS KYFAWT,
       SUM(CASE CODE WHEN 'KYFAHS' THEN 1 ELSE 0 END) AS KYFAHS,
       SUM(CASE CODE WHEN 'JHRWSXD' THEN 1 ELSE 0 END) AS JHRWSXD
FROM TABLE1
GROUP BY ID
ORDER BY ID;

------解决方案--------------------
1、如果code确定可以
select id,
      sum(decode(code, 'KYFAWT', total, 0)) as code_A,
      sum(decode(code, 'KYFAHS', total, 0)) as code_B
      ...
 from (SELECT ID, code, COUNT(1) as total FROM YOUR_TABLE GROUP BY ID, code)；

2、如果不确定 采用动态SQL了