CODEWITHSUNDEEP
phpunix-timestamp
Wheelz
Wheelz

Reputation: 219

Find out how long ago something happened

I've got a webpage which records in UNIX when it was last viewed into a datbase field. I want to pull off a report incuding how many days ago this page was last viewed. I'm doing this in PHP

How do i work this out to display? At the moment I have the field showing when it was last accessed by using gmdate("M D Y ", $UNIXTIMESTAMPFIELD) but instead of the date of last access I want to display something like

23 days ago

Thanks

Kieran

Upvotes: 1

Views: 888

Answers (2)

Sébastien Renauld
Sébastien Renauld

Reputation: 19662

That's pretty easy if you're on PHP5.3 or later. Use the DateInterval object, which can understand unix timestamps and easily output a difference.

What you can do is very, very straightforward. Assume your timestamps are $TS1, $TS2.

Step 1: create the DateTime objects for each:

$DT1 = new DateTime("@{$TS1}");
$DT2 = new DateTime("@{$TS2}");

Step 2: diff them

$diff = $DT1->diff($DT2);

Step 3: print stuff!

echo "Days: ".$diff->d;

This automatically takes into account timezone settings, amongst other things. It also allows you to easily substract from datetime objects should you need to.

Upvotes: 3

Vyktor
Vyktor

Reputation: 20997

Difference is clearly $diff = time() - $UNIXTIMESTAMPFIELD (now-event). Not you just have to format it correctly:

function time_diff_to_human($seconds){
    // For seconds
    if( $seconds < 60){
        if( $seconds == 1){
            return "a second ago"
        }
        return sprintf( "%d seconds ago", $seconds);
    }

    // Minutes
    $minutes = round( $seconds/60);
    if( $minutes < 60){
        if( $minutes == 1){
            return "a minute ago"
        }
        return sprintf( "%d minutes ago", $minutes);
    }

    // Hours
    $hours = round( $minutes/60);
    if( $hours < 24){
        if( $hours == 1){
            return "last hour"
        }
        return sprintf( "%d hours", $hours);
    }

    // Add some formatting to days
    $days = $months/24;
    if( $days < 31){
        if( $days == 1){
            return 'yesterday';
        }
        return sprintf( "%d days", $days);
    }

    // Approx months
    $months = round( $days/30);
    if( $months < 12){
        if( $months == 1){
            return "last month"
        }
        return sprintf( "%d months", $months);
    }

    // And finally years, note that they are calculated from days
    $years = round( $days/365.4);
    if( $years == 1){
        return "last year"
    }
    return sprintf( "%d years", $years);

}

Upvotes: 0

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