请教了两天的问题,老师也不会。MYSQL
表php188_member
值parent_cp_member_active,mem_id,mem_username
要求:如果parent_cp_member_active等于mem_id.则输出mem_username
用了IN、join、自连接。。。
要么空表,要么错误,纠结的头都大了。
那个好心的哥哥姐姐能赐教下。
直接用select子句就行了:
1. 如果直接能找到就不用ifnull判断:
如:
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SQL code
mysql> create table hj89 (mem_id integer primary key, mem_username varchar(32), parent_cp_member_active varchar(32)); Query OK, 0 rows affected (0.17 sec) mysql> insert into hj89 values(32687, '张三', 32457); Query OK, 1 row affected (0.06 sec) mysql> insert into hj89 values(32457, '李四', 32687); Query OK, 1 row affected (0.05 sec) mysql> select mem_id, mem_username, (select mem_username from hj89 where mem_id = c.parent_cp_member_active) from hj89 c; +--------+--------------+--------------------------------------+ | mem_id | mem_username | (select mem_username from hj89 where mem_id = c.parent_cp_member_active) | +--------+--------------+--------------------------------------+ | 32457 | 李四 | 张三 | | 32687 | 张三 | 李四 | +--------+--------------+--------------------------------------+ 2 rows in set (0.03 sec)