一道sql面試題

問題描述：
兩個表a，b
a中欄位為：id,name
b中欄位為:id,email,address
查詢語句：select a.*,b.* from a left join b on a.id=b.id
結果為：

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) 

 問題，如果在left join語句中只取出與附表第一條記錄聯合的結果如何做？即主表資料只和附表第一條聯合，其餘忽略，例如下列結果：
id          name                                               sex                                                id          address                                            email
----------- -------------------------------------------------- -------------------------------------------------- ----------- -------------------------------------------------- ---------------
1           zy                                                 male                                               1           beijing                                            zyip@qq.com
2           zk                                                 male                                               2           beijing                                            zk@qq.com