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
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(', ', $string) . ' ago' : 'just now';
}
其他回答
这将尝试检测是否给定了时间戳,并将返回未来的日期/时间作为负值:
<?php
function time_diff($start, $end = NULL, $convert_to_timestamp = FALSE) {
// If $convert_to_timestamp is not explicitly set to TRUE,
// check to see if it was accidental:
if ($convert_to_timestamp || !is_numeric($start)) {
// If $convert_to_timestamp is TRUE, convert to timestamp:
$timestamp_start = strtotime($start);
}
else {
// Otherwise, leave it as a timestamp:
$timestamp_start = $start;
}
// Same as above, but make sure $end has actually been overridden with a non-null,
// non-empty, non-numeric value:
if (!is_null($end) && (!empty($end) && !is_numeric($end))) {
$timestamp_end = strtotime($end);
}
else {
// If $end is NULL or empty and non-numeric value, assume the end time desired
// is the current time (useful for age, etc):
$timestamp_end = time();
}
// Regardless, set the start and end times to an integer:
$start_time = (int) $timestamp_start;
$end_time = (int) $timestamp_end;
// Assign these values as the params for $then and $now:
$start_time_var = 'start_time';
$end_time_var = 'end_time';
// Use this to determine if the output is positive (time passed) or negative (future):
$pos_neg = 1;
// If the end time is at a later time than the start time, do the opposite:
if ($end_time <= $start_time) {
$start_time_var = 'end_time';
$end_time_var = 'start_time';
$pos_neg = -1;
}
// Convert everything to the proper format, and do some math:
$then = new DateTime(date('Y-m-d H:i:s', $$start_time_var));
$now = new DateTime(date('Y-m-d H:i:s', $$end_time_var));
$years_then = $then->format('Y');
$years_now = $now->format('Y');
$years = $years_now - $years_then;
$months_then = $then->format('m');
$months_now = $now->format('m');
$months = $months_now - $months_then;
$days_then = $then->format('d');
$days_now = $now->format('d');
$days = $days_now - $days_then;
$hours_then = $then->format('H');
$hours_now = $now->format('H');
$hours = $hours_now - $hours_then;
$minutes_then = $then->format('i');
$minutes_now = $now->format('i');
$minutes = $minutes_now - $minutes_then;
$seconds_then = $then->format('s');
$seconds_now = $now->format('s');
$seconds = $seconds_now - $seconds_then;
if ($seconds < 0) {
$minutes -= 1;
$seconds += 60;
}
if ($minutes < 0) {
$hours -= 1;
$minutes += 60;
}
if ($hours < 0) {
$days -= 1;
$hours += 24;
}
$months_last = $months_now - 1;
if ($months_now == 1) {
$years_now -= 1;
$months_last = 12;
}
// "Thirty days hath September, April, June, and November" ;)
if ($months_last == 9 || $months_last == 4 || $months_last == 6 || $months_last == 11) {
$days_last_month = 30;
}
else if ($months_last == 2) {
// Factor in leap years:
if (($years_now % 4) == 0) {
$days_last_month = 29;
}
else {
$days_last_month = 28;
}
}
else {
$days_last_month = 31;
}
if ($days < 0) {
$months -= 1;
$days += $days_last_month;
}
if ($months < 0) {
$years -= 1;
$months += 12;
}
// Finally, multiply each value by either 1 (in which case it will stay the same),
// or by -1 (in which case it will become negative, for future dates).
// Note: 0 * 1 == 0 * -1 == 0
$out = new stdClass;
$out->years = (int) $years * $pos_neg;
$out->months = (int) $months * $pos_neg;
$out->days = (int) $days * $pos_neg;
$out->hours = (int) $hours * $pos_neg;
$out->minutes = (int) $minutes * $pos_neg;
$out->seconds = (int) $seconds * $pos_neg;
return $out;
}
示例用法:
<?php
$birthday = 'June 2, 1971';
$check_age_for_this_date = 'June 3, 1999 8:53pm';
$age = time_diff($birthday, $check_age_for_this_date)->years;
print $age;// 28
Or:
<?php
$christmas_2020 = 'December 25, 2020';
$countdown = time_diff($christmas_2020);
print_r($countdown);
简单的功能
function time_difference($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
您可以使用
getdate()
函数,该函数返回包含所提供日期/时间的所有元素的数组:
$diff = abs($endDate - $startDate);
$my_t=getdate($diff);
print("$my_t[year] years, $my_t[month] months and $my_t[mday] days");
如果开始和结束日期为字符串格式,则使用
$startDate = strtotime($startDateStr);
$endDate = strtotime($endDateStr);
在上述代码之前
我在下面的页面上找到了您的文章,其中包含了许多PHP日期时间计算的参考。
使用PHP计算两个日期(和时间)之间的差异。下一页提供了一系列不同的方法(共7种),用于使用PHP执行日期/时间计算,以确定两个日期之间的时间差(小时、弹药)、天、月或年。
请参阅PHP日期时间–计算两个日期之间差值的7种方法。
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);