infomatica8.5中join transform控件的join type属性

infomatica8.5中join transform控件的join type属性

 

4种: normal = inner join;

         Master Outer Join=RIGHT OUTER JOIN;

         Detail Outer Join=LEFT OUTER JOIN ;

         Full Outer Join=FULL OUTER JOIN.

 

例如:

表A记录如下:
aID               aNum
1                  a20050111
2                  a20050112
3                  a20050113
4                  a20050114
5                  a20050115

表B记录如下:
bID               bName
1                   2006032401
2                  2006032402
3                  2006032403
4                  2006032404
8                  2006032408


实验如下:
1.left join

sql语句如下:
select * from A
left join B
on A.aID = B.bID

结果如下:
aID               aNum                          bID                  bName
1                   a20050111                1                      2006032401
2                   a20050112                2                     2006032402
3                   a20050113                3                     2006032403
4                   a20050114                4                     2006032404
5                   a20050115                NULL              NULL
(所影响的行数为 5 行)

结果说明:
               left join是以A表的记录为基础的,A可以看成左表,B可以看成右表,left join是以左表为准的.
换句话说,左表(A)的记录将会全部表示出来,而右表(B)只会显示符合搜索条件的记录(例子中为: A.aID = B.bID).
B表记录不足的地方均为NULL.

2.right join
sql语句如下:
select * from A
right join B
on A.aID = B.bID
结果如下:
aID               aNum                          bID                  bName
1                   a20050111                1                      2006032401
2                   a20050112                2                     2006032402
3                   a20050113                3                     2006032403
4                   a20050114                4                     2006032404
NULL           NULL                          8                     2006032408
(所影响的行数为 5 行)
结果说明:
        仔细观察一下,就会发现,和left join的结果刚好相反,这次是以右表(B)为基础的,A表不足的地方用NULL填充.


3.inner join
sql语句如下:
select * from A
innerjoin B
on A.aID = B.bID

结果如下:
aID               aNum                          bID                  bName
1                   a20050111                1                      2006032401
2                   a20050112                2                     2006032402
3                   a20050113                3                     2006032403
4                   a20050114                4                     2006032404

结果说明:
        很明显,这里只显示出了 A.aID = B.bID的记录.这说明inner join并不以谁为基础,它只显示符合条件的记录.  

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