mysql date comparison with date_format

2019-01-07 21:56发布

I googled and tried several ways to compare date but unfortunately didn't get the result as expected. I have current state of records like following:

        mysql> select date_format(date(starttime),'%d-%m-%Y') from data;

              +-----------------------------------------+
              | date_format(date(starttime),'%d-%m-%Y') |
              +-----------------------------------------+
              | 28-10-2012                              |
              | 02-11-2012                              |
              | 02-11-2012                              |
              | 02-11-2012                              |
              | 03-11-2012                              |
              | 03-11-2012                              |
              | 07-11-2012                              |
              | 07-11-2012                              |

I would like to compare date and therefore do like this:

        mysql> select date_format(date(starttime),'%d-%m-%Y') from data where date_format(date(starttime),'%d-%m-%y') >= '02-11-2012';
               +-----------------------------------------+
               | date_format(date(starttime),'%d-%m-%Y') |
               +-----------------------------------------+
               | 28-10-2012                              |
               | 02-11-2012                              |
               | 02-11-2012                              |
               | 02-11-2012                              |
               | 03-11-2012                              |
               | 03-11-2012                              |
               | 07-11-2012                              |
               | 07-11-2012                              |

I believe that the result should not include '28-10-2012'. Any suggestion? Thanks in advance.

3条回答
虎瘦雄心在
2楼-- · 2019-01-07 22:38

Use 2012-11-02 instead of 02-11-2012 and you will not need date_format() anymore

查看更多
姐就是有狂的资本
3楼-- · 2019-01-07 22:39

Your format is fundamentally not a sortable one to start with - you're comparing strings, and the string "28-10-2012" is greater than "02-11-2012".

Instead, you should be comparing dates as dates, and then only converting them into your target format for output.

Try this:

select date_format(date(starttime),'%d-%m-%Y') from data
where date(starttime) >= date '2012-11-02';

(The input must always be in year-month-value form, as per the documentation.)

Note that if starttime is a DATETIME field, you might want to consider changing the query to avoid repeated conversion. (The optimizer may well be smart enough to avoid it, but it's worth checking.)

select date_format(date(starttime),'%d-%m-%Y') from data
where starttime >= '2012-11-02 00:00:00';

(Note that it's unusual to format a date as d-m-Y to start with - it would be better to use y-M-d in general, being the ISO-8601 standard etc. However, the above code does what you asked for in the question.)

查看更多
我只想做你的唯一
4楼-- · 2019-01-07 22:45

Use the following method :

public function dateDiff ($date1, $date2) {
/* Return the number of days between the two dates: */
  return round(abs(strtotime($date1)-strtotime($date2))/86400);
}  
/* end function dateDiff */

It will help!

查看更多
登录 后发表回答