0

我有两张桌子

表格1

+-------+-------+
| NAME  | PRICE |
+-------+-------+
| ITEM1 |  100  |
+-------+-------+
| ITEM2 |  200  |
+-------+-------+
| ITEM3 |  300  |
+-------+-------+

table2重复 ITEM1 和 ITEM2

    +-------+--------+---------+
    | NAME  | SUFFIX | CODE    |
    +-------+--------+---------+
    | ITEM1 | 1      | ITEM1_1 |
    +-------+--------+---------+
    | ITEM1 | 2      | ITEM1_2 |
    +-------+--------+---------+
    | ITEM2 | 1      | ITEM2_1 |
    +-------+--------+---------+

我怎样才能用 mySQL 做这个结果？

+-------+-------+----------+
| NAME  | PRICE |   NAME2  |
+-------+-------+----------+
| ITEM1 |  100  |  ITEM1   |
+-------+-------+----------+
| ITEM2 |  200  |  ITEM2   |
+-------+-------+----------+
| ITEM3 |  300  |  NULL    |
+-------+-------+----------+

我想从 table1 中获取不在table2 中的元素。在这种情况下，我想要获取的元素是 ITEM3。我可以用 LEFT JOIN 做到这一点吗？

4

2 回答 2

1 
select * from table1 t1
left outer join table2 t2 on t1.name = t2.name
where t2.name is null
于 2012-06-19T12:48:02.120 回答
1 
Select * from table1 s Left join table2 as t on s.NAME=t.NAME where t.NAME IS NULL
于 2012-06-19T12:48:03.863 回答