一道sql面试题

问题描述：
两个表a，b
a中字段为：id,name
b中字段为:id,email,address
查询语句：select a.*,b.* from a left join b on a.id=b.id
结果为：

id          name                                               sex                                                id          address                                            email
----------- -------------------------------------------------- -------------------------------------------------- ----------- -------------------------------------------------- -
1           zy                                                 male                                               1           beijing                                            zyip@qq.com
1           zy                                                 male                                               1           American                                           zyip@msn.com
2           zk                                                 male                                               2           beijing                                            zk@qq.com
2           zk                                                 male                                               2           nanjing                                            zk@sina.com.cn

(4 row(s) affected) 

 问题，如果在left join语句中只取出与附表第一条记录联合的结果如何做？即主表数据只和附表第一条联合，其余忽略，例如下列结果：
id          name                                               sex                                                id          address                                            email
----------- -------------------------------------------------- -------------------------------------------------- ----------- -------------------------------------------------- ---------------
1           zy                                                 male                                               1           beijing                                            zyip@qq.com
2           zk                                                 male                                               2           beijing                                            zk@qq.com