date / time convert to time since php

Question

Hello in my database date / time are in this format

2010-06-01T18:20:25+0000

I'd like to echo that out to time passed since that date / time

e.g.

4 days 3 hours 36 minutes and 4 seconds

is this possible?

Answer 1

Below is a function I wrote to do this. Feel free to use it.

/**
 * Returns rough (in largest single unit) time elapsed between two times.
 * @param int $iTime0  Initial time, as time_t.
 * @param int $iTime1  Final time, as time_t. 0=use current time.
 * @return string Time elapsed, like "5 minutes" or "3 days" or "1 month".
 *              You might print "ago" after this return if $iTime1 is now.
 * @author Dan Kamins - dos at axonchisel dot net
 */
function ax_getRoughTimeElapsedAsText($iTime0, $iTime1 = 0)
{
    if ($iTime1 == 0) { $iTime1 = time(); }
    $iTimeElapsed = $iTime1 - $iTime0;

    if ($iTimeElapsed < (60)) {
        $iNum = intval($iTimeElapsed); $sUnit = "second";
    } else if ($iTimeElapsed < (60*60)) {
        $iNum = intval($iTimeElapsed / 60); $sUnit = "minute";
    } else if ($iTimeElapsed < (24*60*60)) {
        $iNum = intval($iTimeElapsed / (60*60)); $sUnit = "hour";
    } else if ($iTimeElapsed < (30*24*60*60)) {
        $iNum = intval($iTimeElapsed / (24*60*60)); $sUnit = "day";
    } else if ($iTimeElapsed < (365*24*60*60)) {
        $iNum = intval($iTimeElapsed / (30*24*60*60)); $sUnit = "month";
    } else {
        $iNum = intval($iTimeElapsed / (365*24*60*60)); $sUnit = "year";
    }

    return $iNum . " " . $sUnit . (($iNum != 1) ? "s" : "");
}

To use this func, you'd need to first convert your times to time_t format (integer #seconds since the "epoch"). Either of these PHP functions will probably help with that: http://php.net/strptime or http://php.net/strtotime.

Answer 2

I changed one line of the above code for the case in which it was posted less than one minute ago.

    function ax_getRoughTimeElapsedAsText($iTime0, $iTime1 = 0)
    {
    if ($iTime1 == 0) { $iTime1 = time(); }
    $iTimeElapsed = $iTime1 - $iTime0;

    if ($iTimeElapsed < (60)) {
        return "Less than a minute ago";
    } else if ($iTimeElapsed < (60*60)) {
        $iNum = intval($iTimeElapsed / 60); $sUnit = "minute";
    } else if ($iTimeElapsed < (24*60*60)) {
        $iNum = intval($iTimeElapsed / (60*60)); $sUnit = "hour";
    } else if ($iTimeElapsed < (30*24*60*60)) {
        $iNum = intval($iTimeElapsed / (24*60*60)); $sUnit = "day";
    } else if ($iTimeElapsed < (365*24*60*60)) {
        $iNum = intval($iTimeElapsed / (30*24*60*60)); $sUnit = "month";
    } else {
        $iNum = intval($iTimeElapsed / (365*24*60*60)); $sUnit = "year";
    }

    return $iNum . " " . $sUnit . (($iNum != 1) ? "s" : "") . " ago";
    }

Answer 3

you can use the func timediff right in your database:

http://dev.mysql.com/doc/refman/5.1/en/date-and-time-functions.html#function_timediff

pass the first param as the date, and the second param as now()

Answer 4

function time_ago($timestamp, $granularity = 2) {
  $timestamp = time() - $timestamp;
  $units = array('1 year|%d years' => 31536000, 
                 '1 week|%d weeks' => 604800, 
                 '1 day|%d days' => 86400, 
                 '1 hour|%d hours' => 3600, 
                 '1 min|%d mins' => 60, 
                 '1 sec|%d secs' => 1
                );
  $output = '';
  foreach ($units as $key => $value) {
    $key = explode('|', $key);
    if ($timestamp >= $value) {
      $pluralized = floor($timestamp / $value) > 1 ? 
                    sprintf($key[1], floor($timestamp / $value)) : 
                    $key[0];
      $output .= ($output ? ' ' : '') . $pluralized;
      $timestamp %= $value;
      $granularity--;
    }
    if ($granularity == 0) {
      break;
    }
  }
  return $output ? $output : "Just now";
}

This should be close.

Edit: added this line: $timestamp = time() - $timestamp;

Answer 5

u can fetch that fron db and use strtotime function which will make into epoch . and then use date command to print it into any format u want to .