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
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);
您还可以使用以下代码通过向上舍入分数来返回日期差异$date1=$duedate;//指定到期日echo$date2=日期(“Y-m-d”);//当前日期$ts1=字符串时间($date1);$ts2=字符串时间($date2);$seconds_diff=$ts1-$ts2;echo$datediff=ceil(($seconds_diff/3600)/24);//天内返回
如果您使用php的floor方法而不是ceil,它将返回舍入分数。请检查此处的差异,有时,如果您的临时服务器时区与现场站点时区不同,在这种情况下,您可能会得到不同的结果,因此请相应地更改条件。
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
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);
// If you just want to see the year difference then use this function.
// Using the logic I've created you may also create month and day difference
// which I did not provide here so you may have the efforts to use your brain.
// :)
$date1='2009-01-01';
$date2='2010-01-01';
echo getYearDifference ($date1,$date2);
function getYearDifference($date1=strtotime($date1),$date2=strtotime($date2)){
$year = 0;
while($date2 > $date1 = strtotime('+1 year', $date1)){
++$year;
}
return $year;
}