PHP: date “Yesterday”, “Today”

Question

I have a little function that shows latest activity, it grab timestamp in unix format from the db, and then it echo out with this line:

 date(\"G:i:s j M -Y\         


        

Answer 1

here's working function

    function minus_one_day($date){
      $date2 = formatDate4db($date);
      $date1 = str_replace('-', '/', $date2);
      $yesterday = date('Y-m-d',strtotime($date1 . "-1 days"));
      return $yesterday; }

hope work for you...

Answer 2

I would find the timestap for last midnight and the one before it, if $last_access is between the two timestamps, then display yesterday, anything greater than last midnight's timestamp would be today...

I believe that would be the quicker than doing date arithmetic.

Actually, I just tested this code and it seems to work great:

<?php
    if ($last_access >= strtotime("today"))
        echo "Today";
    else if ($last_access >= strtotime("yesterday"))
        echo "Yesterday";
?>

Answer 3

I enhanced Thomas Decaux answer to come up with this

function formatTimeString($timeStamp) {
$str_time = date("Y-m-d H:i:sP", $timeStamp);
$time = strtotime($str_time);
$d = new DateTime($str_time);

$weekDays = ['Mon', 'Tue', 'Wed', 'Thur', 'Fri', 'Sat', 'Sun'];
$months = ['Jan', 'Feb', 'Mar', 'Apr', ' May', 'Jun', 'Jul', 'Aug', 'Sept', 'Oct', 'Nov', 'Dec'];

if ($time > strtotime('-2 minutes')) {
  return 'Just now';
} elseif ($time > strtotime('-59 minutes')) {
  $min_diff = floor((strtotime('now') - $time) / 60);
  return $min_diff . ' min' . (($min_diff != 1) ? "s" : "") . ' ago';
} elseif ($time > strtotime('-23 hours')) {
  $hour_diff = floor((strtotime('now') - $time) / (60 * 60));
  return $hour_diff . ' hour' . (($hour_diff != 1) ? "s" : "") . ' ago';
} elseif ($time > strtotime('today')) {
  return $d->format('G:i');
} elseif ($time > strtotime('yesterday')) {
  return 'Yesterday at ' . $d->format('G:i');
} elseif ($time > strtotime('this week')) {
  return $weekDays[$d->format('N') - 1] . ' at ' . $d->format('G:i');
} else {
  return $d->format('j') . ' ' . $months[$d->format('n') - 1] .
  (($d->format('Y') != date("Y")) ? $d->format(' Y') : "") .
  ' at ' . $d->format('G:i');
}

}

It takes in the time stamp as the argument, it adds the year of the time if it was from a different year, etc...