我有两个表格日期:
Start Date: 2007-03-24
End Date: 2009-06-26
现在我需要通过以下形式找到这两者之间的区别:
2 years, 3 months and 2 days
如何在PHP中执行此操作?
我有两个表格日期:
Start Date: 2007-03-24
End Date: 2009-06-26
现在我需要通过以下形式找到这两者之间的区别:
2 years, 3 months and 2 days
如何在PHP中执行此操作?
当前回答
这是我的职责。所需PHP>=5.3.4。它使用DateTime类。非常快,很快,可以区分两个日期,甚至所谓的“开始时间”。
if(function_exists('grk_Datetime_Since') === FALSE){
function grk_Datetime_Since($From, $To='', $Prefix='', $Suffix=' ago', $Words=array()){
# Est-ce qu'on calcul jusqu'à un moment précis ? Probablement pas, on utilise maintenant
if(empty($To) === TRUE){
$To = time();
}
# On va s'assurer que $From est numérique
if(is_int($From) === FALSE){
$From = strtotime($From);
};
# On va s'assurer que $To est numérique
if(is_int($To) === FALSE){
$To = strtotime($To);
}
# On a une erreur ?
if($From === FALSE OR $From === -1 OR $To === FALSE OR $To === -1){
return FALSE;
}
# On va créer deux objets de date
$From = new DateTime(@date('Y-m-d H:i:s', $From), new DateTimeZone('GMT'));
$To = new DateTime(@date('Y-m-d H:i:s', $To), new DateTimeZone('GMT'));
# On va calculer la différence entre $From et $To
if(($Diff = $From->diff($To)) === FALSE){
return FALSE;
}
# On va merger le tableau des noms (par défaut, anglais)
$Words = array_merge(array(
'year' => 'year',
'years' => 'years',
'month' => 'month',
'months' => 'months',
'week' => 'week',
'weeks' => 'weeks',
'day' => 'day',
'days' => 'days',
'hour' => 'hour',
'hours' => 'hours',
'minute' => 'minute',
'minutes' => 'minutes',
'second' => 'second',
'seconds' => 'seconds'
), $Words);
# On va créer la chaîne maintenant
if($Diff->y > 1){
$Text = $Diff->y.' '.$Words['years'];
} elseif($Diff->y == 1){
$Text = '1 '.$Words['year'];
} elseif($Diff->m > 1){
$Text = $Diff->m.' '.$Words['months'];
} elseif($Diff->m == 1){
$Text = '1 '.$Words['month'];
} elseif($Diff->d > 7){
$Text = ceil($Diff->d/7).' '.$Words['weeks'];
} elseif($Diff->d == 7){
$Text = '1 '.$Words['week'];
} elseif($Diff->d > 1){
$Text = $Diff->d.' '.$Words['days'];
} elseif($Diff->d == 1){
$Text = '1 '.$Words['day'];
} elseif($Diff->h > 1){
$Text = $Diff->h.' '.$Words['hours'];
} elseif($Diff->h == 1){
$Text = '1 '.$Words['hour'];
} elseif($Diff->i > 1){
$Text = $Diff->i.' '.$Words['minutes'];
} elseif($Diff->i == 1){
$Text = '1 '.$Words['minute'];
} elseif($Diff->s > 1){
$Text = $Diff->s.' '.$Words['seconds'];
} else {
$Text = '1 '.$Words['second'];
}
return $Prefix.$Text.$Suffix;
}
}
其他回答
简单的功能
function time_difference($time_1, $time_2, $limit = null)
{
$val_1 = new DateTime($time_1);
$val_2 = new DateTime($time_2);
$interval = $val_1->diff($val_2);
$output = array(
"year" => $interval->y,
"month" => $interval->m,
"day" => $interval->d,
"hour" => $interval->h,
"minute" => $interval->i,
"second" => $interval->s
);
$return = "";
foreach ($output AS $key => $value) {
if ($value == 1)
$return .= $value . " " . $key . " ";
elseif ($value >= 1)
$return .= $value . " " . $key . "s ";
if ($key == $limit)
return trim($return);
}
return trim($return);
}
像这样使用
回波时间差($time_1,$time_2,“天”);
将返回2年8个月2天
我建议使用DateTime和DateInterval对象。
$date1 = new DateTime("2007-03-24");
$date2 = new DateTime("2009-06-26");
$interval = $date1->diff($date2);
echo "difference " . $interval->y . " years, " . $interval->m." months, ".$interval->d." days ";
// shows the total amount of days (not divided into years, months and days like above)
echo "difference " . $interval->days . " days ";
阅读更多php DateTime::diff手册
根据手册:
从PHP 5.2.2开始,DateTime对象可以使用比较运算符进行比较。
$date1 = new DateTime("now");
$date2 = new DateTime("tomorrow");
var_dump($date1 == $date2); // bool(false)
var_dump($date1 < $date2); // bool(true)
var_dump($date1 > $date2); // bool(false)
我在PHP5.2中遇到了同样的问题,并用MySQL解决了这个问题。可能并不是你想要的,但这会奏效,并返回天数:
$datediff_q = $dbh->prepare("SELECT DATEDIFF(:date2, :date1)");
$datediff_q->bindValue(':date1', '2007-03-24', PDO::PARAM_STR);
$datediff_q->bindValue(':date2', '2009-06-26', PDO::PARAM_STR);
$datediff = ($datediff_q->execute()) ? $datediff_q->fetchColumn(0) : false;
此处有更多信息http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html#function_datediff
由于每个人都在发布代码示例,这里有另一个版本。
我想要一个函数来显示从秒到年的差异(仅一个单位)。对于超过1天的时段,我希望它在午夜滚动(周一上午10点到周三上午9点是2天前,而不是1天前)。对于超过一个月的时间段,我希望滚动在当月的同一天(包括30/31天的月份和闰年)。
这就是我想到的:
/**
* Returns how long ago something happened in the past, showing it
* as n seconds / minutes / hours / days / weeks / months / years ago.
*
* For periods over a day, it rolls over at midnight (so doesn't depend
* on current time of day), and it correctly accounts for month-lengths
* and leap-years (months and years rollover on current day of month).
*
* $param string $timestamp in DateTime format
* $return string description of interval
*/
function ago($timestamp)
{
$then = date_create($timestamp);
// for anything over 1 day, make it rollover on midnight
$today = date_create('tomorrow'); // ie end of today
$diff = date_diff($then, $today);
if ($diff->y > 0) return $diff->y.' year'.($diff->y>1?'s':'').' ago';
if ($diff->m > 0) return $diff->m.' month'.($diff->m>1?'s':'').' ago';
$diffW = floor($diff->d / 7);
if ($diffW > 0) return $diffW.' week'.($diffW>1?'s':'').' ago';
if ($diff->d > 1) return $diff->d.' day'.($diff->d>1?'s':'').' ago';
// for anything less than 1 day, base it off 'now'
$now = date_create();
$diff = date_diff($then, $now);
if ($diff->d > 0) return 'yesterday';
if ($diff->h > 0) return $diff->h.' hour'.($diff->h>1?'s':'').' ago';
if ($diff->i > 0) return $diff->i.' minute'.($diff->i>1?'s':'').' ago';
return $diff->s.' second'.($diff->s==1?'':'s').' ago';
}
使用示例:
echo time_diff_string('2013-05-01 00:22:35', 'now');
echo time_diff_string('2013-05-01 00:22:35', 'now', true);
输出:
4 months ago
4 months, 2 weeks, 3 days, 1 hour, 49 minutes, 15 seconds ago
功能:
function time_diff_string($from, $to, $full = false) {
$from = new DateTime($from);
$to = new DateTime($to);
$diff = $to->diff($from);
$diff->w = floor($diff->d / 7);
$diff->d -= $diff->w * 7;
$string = array(
'y' => 'year',
'm' => 'month',
'w' => 'week',
'd' => 'day',
'h' => 'hour',
'i' => 'minute',
's' => 'second',
);
foreach ($string as $k => &$v) {
if ($diff->$k) {
$v = $diff->$k . ' ' . $v . ($diff->$k > 1 ? 's' : '');
} else {
unset($string[$k]);
}
}
if (!$full) $string = array_slice($string, 0, 1);
return $string ? implode(', ', $string) . ' ago' : 'just now';
}