oracle 更新相同表问题

描述：ta,tb两表的结构完全相同，现在想要以tb中的数据去更新ta表，
要求：以ta为准，若ta中没有的数据，将tb中的数据完全合并到ta中；
若ta中有的数据，但不完全，一些字段为空，那么将tb中相同id的字段去更新ta表，
--
方法一:用全连接，结合nvl函数：
with ta as(
select 1 id, 23 age, 'lilei' name, 'ddd@126.com' mail from dual union all
select 2, null, 'hanmeimei',null from dual union all
select 3, 23, null, 'jim eee@153.com' from dual union all
select 4, 22, 'tom',null from dual),
tb as(
select 1 id, 23 age, 'lilei' name, 'bbb@126.com' mail from dual union all
select 2, 25, 'hanmeimei', 'fff@124com' from dual union all
select 5, 27, 'green', 'ejorj@125.com' from dual)
select nvl(ta.id,tb.id) id,
nvl(ta.age,tb.age) age,
nvl(ta.name,tb.name) name,
nvl(ta.mail,tb.mail) mail
from ta full join tb
on ta.id=tb.id
order by id;
ID        AGE NAME      MAIL
---------- ---------- --------- ---------------
1         23 lilei     ddd@126.com
2         25 hanmeimei fff@124com
3         23           jim eee@153.com
4         22 tom
5         27 green     ejorj@125.com
--
方法二:使用merge into合并：
create table ta(id varchar2(2),age number(3),name varchar2(10),mail varchar2(30));
select * from ta;
ID  AGE NAME       MAIL
-- ---- ---------- ------------------------------
1    23 lilei      ddd@126.com
2       hanmeimei
3    23 jim        eee@153.com
4    22 tom
--
create table tb as select * from ta where 1=0;
select * from tb;
ID  AGE NAME       MAIL
-- ---- ---------- ------------------------------
1    23 lilei      bbb@126.com
2    25 hanmeimei  fff@124.com
5    27 green      ejorj@125.com
--
merge into ta
using tb on (ta.id=tb.id)
when matched then
update set
age=COALESCE(ta.age,tb.age),
name=COALESCE(ta.name,tb.name),
mail=COALESCE(ta.mail,tb.mail)
when not matched then
insert(ta.id,ta.age,ta.name,ta.mail)
values(tb.id,tb.age,tb.name,tb.mail);
--
ID  AGE NAME       MAIL
-- ---- ---------- ------------------------------
1    23 lilei      ddd@126.com
2    25 hanmeimei  fff@124.com
3    23 jim        eee@153.com
4    22 tom
5    27 green      ejorj@125.com
--
方法三:使用update直接更新ta表,若ta中没有的数据,将tb中的数据添加进来即可：
3.1 更新
update ta a
set (a.age,a.name,a.mail)=(
select nvl(a1.age,b1.age),
nvl(a1.name,b1.name),
nvl(a1.mail,b1.mail)
from ta a1,tb b1
where a1.id=b1.id and a1.id=a.id)
where exists (select 1 from ta a2 where a2.id=a.id);
//注意：此方法失败，如果您能想到解决办法，请赐教
3.2 添加
insert into ta(id,age,name,mail)
select tb.id,tb.age,tb.name,tb.mail
from tb
where tb.id not in(select ta.id from ta);