<delect id="8ebfo"><tfoot id="8ebfo"><strike id="8ebfo"></strike></tfoot></delect>
<tr id="8ebfo"></tr>

文章內(nèi)容

在Oracle中，單引號有兩個作用:

發(fā)布時間: 2012/9/12 17:27:48

描述：ta,tb兩表的結(jié)構(gòu)完全相同，現(xiàn)在想要以tb中的數(shù)據(jù)去更新ta表，
要求：以ta為準(zhǔn)，若ta中沒有的數(shù)據(jù)，將tb中的數(shù)據(jù)完全合并到ta中；
      若ta中有的數(shù)據(jù)，但不完全，一些字段為空，那么將tb中相同id的字段去更新ta表，
--
方法一:用全連接，結(jié)合nvl函數(shù)：
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中沒有的數(shù)據(jù),將tb中的數(shù)據(jù)添加進(jìn)來即可：
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：此方法失敗，如果您能想到解決辦法，請賜教
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);

本文出自：億恩科技【www.artduck.net】

服務(wù)器租用/服務(wù)器托管中國五強！虛擬主機域名注冊頂級提供商！15年品質(zhì)保障！--億恩科技[ENKJ.COM]