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
对于php版本>=5.3:创建两个日期对象,然后使用date_diff()函数。它将返回phpDateInterval对象。请参阅文档
$date1=date_create("2007-03-24");
$date2=date_create("2009-06-26");
$diff=date_diff($date1,$date2);
echo $diff->format("%R%a days");
其他回答
使用示例:
echo time_diff_string('2013-05-01 00:22:35', 'now');
echo time_diff_string('2013-05-01 00:22:35', 'now', true);
输出:
4 months ago
4 months, 2 weeks, 3 days, 1 hour, 49 minutes, 15 seconds ago
功能:
function time_diff_string($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';
}
这是可运行的代码
$date1 = date_create('2007-03-24');
$date2 = date_create('2009-06-26');
$diff1 = date_diff($date1,$date2);
$daysdiff = $diff1->format("%R%a");
$daysdiff = abs($daysdiff);
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()) ? $datediff_q->fetchColumn(0) : false;
此处有更多信息http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html#function_datediff
当PHP 5.3(分别为date_diff())不可用时,我使用了我编写的以下函数:
function dateDifference($startDate, $endDate)
{
$startDate = strtotime($startDate);
$endDate = strtotime($endDate);
if ($startDate === false || $startDate < 0 || $endDate === false || $endDate < 0 || $startDate > $endDate)
return false;
$years = date('Y', $endDate) - date('Y', $startDate);
$endMonth = date('m', $endDate);
$startMonth = date('m', $startDate);
// Calculate months
$months = $endMonth - $startMonth;
if ($months <= 0) {
$months += 12;
$years--;
}
if ($years < 0)
return false;
// Calculate the days
$measure = ($months == 1) ? 'month' : 'months';
$days = $endDate - strtotime('+' . $months . ' ' . $measure, $startDate);
$days = date('z', $days);
return array($years, $months, $days);
}