帐号表结构如下： CREATE TABLE `t_account` ( `account_id` int(10) unsigned NOT NULL auto_increment COMMENT '所属帐号id', `account_name` varchar(50) NOT NULL COMMENT '帐号名', `password` varchar(16) NOT NULL COMMENT '密码', `last_login_time` timestamp NOT NULL default '1979-01-01 00:00:00' COMMENT '最后登录时间', `point` int(10) unsigned NOT NULL default '0' COMMENT '积分', PRIMARY KEY USING BTREE (`account_id`), ) ENGINE=InnoDB AUTO_INCREMENT=10001 DEFAULT CHARSET=utf8; CREATE TABLE `t_account_nickname` ( `account_id` int(10) unsigned NOT NULL COMMENT '所属帐号id', `account_nickname `varchar(50) NOT NULL COMMENT'昵称', PRIMARY KEY USING BTREE (`account_id`), ) ENGINE=InnoDB DEFAULT CHARSET=utf8; 请写出SQL语句（温馨提示：列点作答，字体端正，代码细节与规范） 1.查询最近1000个登录的帐号（mysql） 2.查询如下结果：第一列帐号ID，第二列帐号名，第三列昵称，第四列积分 3.对帐号名里包含rock的帐号，积分修改为0 4.查询出重复的昵称

1. 查询最近1000个登录的帐号（mysql）

SELECT account_id, account_name, last_login_time, point 
FROM t_account 
ORDER BY last_login_time DESC 
LIMIT 1000;

2. 查询如下结果：第一列帐号ID，第二列帐号名，第三列昵称，第四列积分

SELECT t1.account_id, t1.account_name, t2.account_nickname, t1.point 
FROM t_account AS t1 
LEFT JOIN t_account_nickname AS t2 ON t1.account_id = t2.account_id;

3. 对帐号名里包含rock的帐号，积分修改为0

UPDATE t_account 
SET point = 0 
WHERE account_name LIKE '%rock%';

4. 查询出重复的昵称

SELECT account_nickname, COUNT(*) AS num 
FROM t_account_nickname 
GROUP BY account_nickname 
HAVING num > 1;