How can i create (before 1 day, 2 day, month, year ...)

[web_designer_iq]

Hi
I have this code:

$tStamp = intval((intval($dateUnixTimeStamp) - date("U")) / 86400);

this code to get the timestamp days.
i want to create it to calculate the day(s) & month(s) & year (s)
For example:
Before 1 week
Before 1 month
Before 1 month & 5 day
Before 1 month & 1 week
etc...
How can i create it please?

[devinemke]

have a look at [man]strtotime[/man]

echo 'today: ' . date('Y-m-d') . '<br>';
echo '1 week ago: ' . date('Y-m-d', strtotime('-1 week')) . '<br>';
echo '1 month ago: ' . date('Y-m-d', strtotime('-1 month')) . '<br>';
echo '1 month and 5 days ago: ' . date('Y-m-d', strtotime('-1 month -5 days')) . '<br>';
echo '1 month and 1 week ago: ' . date('Y-m-d', strtotime('-1 month -1 week')) . '<br>';

[web_designer_iq]

devinemke

Thank you :-)
But it was just an example ! I want to calculate every time.

[devinemke]

web.designer.iq;10913606 wrote:

I want to calculate every time.

then i don't understand the question

[web_designer_iq]

OK
Let me to explain it.
You have a timestamp and you need to calculate the days and order them (the days) by (Days, Weeks, Months, Years).
For example:
I have an articles it's created in 1240921974 timestamp. How can i type "This article create since 1 week". or this article is created since 1 month and 3 days.

[coldwerturkey]

here's a function i wrote a while ago,

function timeago($time) {
	//Find difference in seconds
    $d = time()-$time;
    //Second(s)
    if ($d < 60) {
        return $d." second".(($d==1)?'':'s')." ago";
    } else {
        //Minute(s)
        $d = round($d / 60);
        if($d < 60) {
            return $d." minute".(($d==1)?'':'s')." ago";
        } else {
            //Hour(s)
            $d = round($d / 60);
            if($d < 24) {
                return $d." hour".(($d==1)?'':'s')." ago";
            } else {
                //Day(s)
                $d = round($d / 24);
                if($d < 7) {
                    return $d." day".(($d==1)?'':'s')." ago";
                } else {
                    //Week(s)
                    $d = round($d / 7);
                    if($d < 1) {
                        return $d." week ago";
                    }
                }
            }
        }
    }
}

here's a another one from the comments in time() from php.net

public static function distanceOfTimeInWords($fromTime, $toTime = 0, $showLessThanAMinute = false) {
    $distanceInSeconds = round(abs($toTime - $fromTime));
    $distanceInMinutes = round($distanceInSeconds / 60);

    if ( $distanceInMinutes <= 1 ) {
        if ( !$showLessThanAMinute ) {
            return ($distanceInMinutes == 0) ? 'less than a minute' : '1 minute';
        } else {
            if ( $distanceInSeconds < 5 ) {
                return 'less than 5 seconds';
            }
            if ( $distanceInSeconds < 10 ) {
                return 'less than 10 seconds';
            }
            if ( $distanceInSeconds < 20 ) {
                return 'less than 20 seconds';
            }
            if ( $distanceInSeconds < 40 ) {
                return 'about half a minute';
            }
            if ( $distanceInSeconds < 60 ) {
                return 'less than a minute';
            }

            return '1 minute';
        }
    }
    if ( $distanceInMinutes < 45 ) {
        return $distanceInMinutes . ' minutes';
    }
    if ( $distanceInMinutes < 90 ) {
        return 'about 1 hour';
    }
    if ( $distanceInMinutes < 1440 ) {
        return 'about ' . round(floatval($distanceInMinutes) / 60.0) . ' hours';
    }
    if ( $distanceInMinutes < 2880 ) {
        return '1 day';
    }
    if ( $distanceInMinutes < 43200 ) {
        return 'about ' . round(floatval($distanceInMinutes) / 1440) . ' days';
    }
    if ( $distanceInMinutes < 86400 ) {
        return 'about 1 month';
    }
    if ( $distanceInMinutes < 525600 ) {
        return round(floatval($distanceInMinutes) / 43200) . ' months';
    }
    if ( $distanceInMinutes < 1051199 ) {
        return 'about 1 year';
    }

    return 'over ' . round(floatval($distanceInMinutes) / 525600) . ' years';
}