5 min ago in php

Question

I have stored a date in database as "15-10-2012",while showing in the front end when the date is not older than one day.I need to show 2 min age , 5 hours ago etc.

how can i achieve this? thanks in advance.

Input

15-10-2012

Output

"2 days ago"

[calculating and showing a date as 'secs ago', 'mins ago', 'hours ago' etc](http://stackoverflow.com/q/2452010) — Pekka, Oct 16 '12 at 10:12
[Calculate time difference between two dates, and present the answer like "2 days 3 hours ago"](http://stackoverflow.com/q/432398) — Pekka, Oct 16 '12 at 10:12
possible duplicate of [Calculate time difference between two dates, and present the answer like "2 days 3 hours ago"](http://stackoverflow.com/questions/432398/calculate-time-difference-between-two-dates-and-present-the-answer-like-2-days) — MrCode, Oct 16 '12 at 10:15

Bogdan Burym · Accepted Answer · 2012-10-16T10:33:48.613

1

See this article http://www.ozzu.com/programming-forum/mysql-selecting-timestamp-from-database-now-minute-t82112.html

And take a look at MYSQL's INTERVAL http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html

I think this will help you: WHERE UNIX_TIMESTAMP(your_date) > (NOW() - INTERVAL 5 MINUTE)

edited Oct 16 '12 at 10:33

answered Oct 16 '12 at 10:27

Bogdan Burym

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Maxim Krizhanovsky · Answer 2 · 2012-10-16T10:23:54.277

0

First calculate a unix timestamp from the date, then substract from current time and turn the difference in minutes/hours/days.

list($day,$month,$year) = explode('-', $date);
$time = mktime(0,0,0,$month,$day,$year);
$now = time();
$difference = $now - $time;
$minutes = intval($difference / 60);

$hours = intval($minutes / 60 ); 

if ($hours > 24) {
   $days = intval($hours / 24);
}

edited Oct 16 '12 at 10:23

answered Oct 16 '12 at 10:10

Maxim Krizhanovsky

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But this won't do what he wants, will it? – Pekka Oct 16 '12 at 10:21

score 0 · Answer 3 · answered Oct 16 '12 at 10:31

(assuming day-month-year)

2 minutes ago:

echo date('j-n-Y', strtotime('15-10-2012') - 120);

Or:

echo date('j-n-Y', strtotime('15-10-2012 -2 minutes'));

Really your starting date should be a timestamp down to the second. If storing just the day/month/year then you'd need to assume each date starts at 00:01

5 min ago in php

3 Answers3