现有oracle表 A 表B
ID COLA ID COLB
1 1 2 4
2 2 3 5
3 3 4 6
对表A和表B全连接FULL JOIN 操作 SQL1：SELECT T.ID,COLA,COLB FROM (SELECT * FROM A) T FULL JOIN (SELECT * FROM B) R ON T.ID=R.ID
结果是 ID COLA COLB
1 1
2 2 4
3 3 5
6 由于SQL1里选择的是T.ID(就是A的ID)，因此全连接后连接字段只会显示A的ID,而B的ID是null
同理如果选择R.ID(就是B的ID)结果如下 A的ID就是null SQL2：SELECT R.ID,COLA,COLB FROM (SELECT * FROM A) T FULL JOIN (SELECT * FROM B) R ON T.ID=R.ID
ID COLA COLB
1
2 2 4
3 3 5
4 6
对结果表处理下即可 SQL3 : SELECT A||B AS ID,COLA,COLB FROM (SELECT T.ID AS A,R.ID AS B,COLA,COLB FROM (SELECT * FROM A) T FULL JOIN (SELECT * FROM B) R ON T.ID=R.ID) WHERE A IS NULL OR B IS NULL UNION ALL SELECT A AS ID,COLA,COLB FROM (SELECT T.ID AS A,R.ID AS B,COLA,COLB FROM (SELECT * FROM A) T FULL JOIN (SELECT * FROM B) R ON T.ID=R.ID) WHERE A IS NOT NULL AND B IS NOT NULL
得到将表A 表B 里的ID取别名 A B ，然后把AB列融合成1列ID,即AB同时不为空的那么取其中一个即可，union 上AB有一个为空的，取A||B即可
A B COLA COLB
1 1
2 2 2 4
3 3 3 5
4 6
最后得到
ID COLA COLB
1 1
2 2 4
3 3 5
4 6