How to SELECT by MAX(date)?

问题:

This is the table structure

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

I need a SELECT statement that will list the report_id per computer_id from latest entered date_entered, and i have no clue how to do that. Can anyone point me into the right direction? Thx in advance.

回答1:

This should do it:

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
)

回答2:

Are you only wanting it to show the last date_entered, or to order by starting with the last_date entered?

SELECT report_id, computer_id, date_entered
FROM reports
GROUP BY computer_id
ORDER BY date_entered DESC
-- LIMIT 1 -- uncomment to only show the last date.

回答3:

Accordig to this: https://bugs.mysql.com/bug.php?id=54784 casting as char should do the trick:

SELECT report_id, computer_id, MAX(CAST(date_entered AS CHAR))
FROM reports
GROUP BY report_id, computer_id

回答4:

This is a very old question but I came here due to the same issue, so I am leaving this here to help any others.

I was trying to optimize the query because it was taking over 5 minutes to query the DB due to the amount of data. My query was similar to the accepted answer's query. Pablo's comment pushed me in the right direction and my 5 minute query became 0.016 seconds. So to help any others that are having very long query times try using an uncorrelated subquery.

The example for the OP would be:

SELECT 
    a.report_id, 
    a.computer_id, 
    a.date_entered
FROM reports AS a
    JOIN (
        SELECT report_id, computer_id, MAX(date_entered) as max_date_entered
        FROM reports
        GROUP BY report_id, computer_id
    ) as b
WHERE a.report_id = b.report_id
    AND a.computer_id = b.computer_id
    AND a.date_entered = b.max_date_entered

Thank you Pablo for the comment. You saved me big time!

回答5:

SELECT report_id, computer_id, date_entered
FROM reports
WHERE date_entered = (
    SELECT date_entered 
    FROM reports 
    ORDER date_entered 
    DESC LIMIT 1
)