如何通过MAX（日期）选择？

Question

这是表结构

CREATE TABLE `reports` (
  `report_id` int(11) NOT NULL auto_increment,
  `computer_id` int(11) NOT NULL default '0',
  `date_entered` datetime NOT NULL default '1970-01-01 00:00:00',
  `total_seconds` int(11) NOT NULL default '0',
  `iphone_id` int(11) default '0',
  PRIMARY KEY  (`report_id`),
  KEY `computer_id` (`computer_id`),
  KEY `iphone_id` (`iphone_id`)
) ENGINE=MyISAM AUTO_INCREMENT=120990 DEFAULT CHARSET=latin1

我需要一条SELECT声明，列出从最近输入的report_id每条消息，我不知道如何执行此操作。谁能指出我正确的方向？提前谢谢。computer_id``date_entered

Answer 1

应该这样做：

SELECT report_id, computer_id, date_entered
FROM reports AS a
WHERE date_entered = (
    SELECT MAX(date_entered)
    FROM reports AS b
    WHERE a.report_id = b.report_id
      AND a.computer_id = b.computer_id
)