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
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,$time_2,“天”);
将返回2年8个月2天
其他回答
我有一些简单的逻辑:
<?php
per_days_diff('2011-12-12','2011-12-29')
function per_days_diff($start_date, $end_date) {
$per_days = 0;
$noOfWeek = 0;
$noOfWeekEnd = 0;
$highSeason=array("7", "8");
$current_date = strtotime($start_date);
$current_date += (24 * 3600);
$end_date = strtotime($end_date);
$seassion = (in_array(date('m', $current_date), $highSeason))?"2":"1";
$noOfdays = array('');
while ($current_date <= $end_date) {
if ($current_date <= $end_date) {
$date = date('N', $current_date);
array_push($noOfdays,$date);
$current_date = strtotime('+1 day', $current_date);
}
}
$finalDays = array_shift($noOfdays);
//print_r($noOfdays);
$weekFirst = array("week"=>array(),"weekEnd"=>array());
for($i = 0; $i < count($noOfdays); $i++)
{
if ($noOfdays[$i] == 1)
{
//echo "This is week";
//echo "<br/>";
if($noOfdays[$i+6]==7)
{
$noOfWeek++;
$i=$i+6;
}
else
{
$per_days++;
}
//array_push($weekFirst["week"],$day);
}
else if($noOfdays[$i]==5)
{
//echo "This is weekend";
//echo "<br/>";
if($noOfdays[$i+2] ==7)
{
$noOfWeekEnd++;
$i = $i+2;
}
else
{
$per_days++;
}
//echo "After weekend value:- ".$i;
//echo "<br/>";
}
else
{
$per_days++;
}
}
/*echo $noOfWeek;
echo "<br/>";
echo $noOfWeekEnd;
echo "<br/>";
print_r($per_days);
echo "<br/>";
print_r($weekFirst);
*/
$duration = array("weeks"=>$noOfWeek, "weekends"=>$noOfWeekEnd, "perDay"=>$per_days, "seassion"=>$seassion);
return $duration;
?>
“如果”日期存储在MySQL中,我发现在数据库级别进行差异计算更容易。。。然后根据“天”、“小时”、“分钟”、“秒”输出,分析并显示相应的结果。。。
mysql> select firstName, convert_tz(loginDate, '+00:00', '-04:00') as loginDate, TIMESTAMPDIFF(DAY, loginDate, now()) as 'Day', TIMESTAMPDIFF(HOUR, loginDate, now())+4 as 'Hour', TIMESTAMPDIFF(MINUTE, loginDate, now())+(60*4) as 'Min', TIMESTAMPDIFF(SECOND, loginDate, now())+(60*60*4) as 'Sec' from User_ where userId != '10158' AND userId != '10198' group by emailAddress order by loginDate desc;
+-----------+---------------------+------+------+------+--------+
| firstName | loginDate | Day | Hour | Min | Sec |
+-----------+---------------------+------+------+------+--------+
| Peter | 2014-03-30 18:54:40 | 0 | 4 | 244 | 14644 |
| Keith | 2014-03-30 18:54:11 | 0 | 4 | 244 | 14673 |
| Andres | 2014-03-28 09:20:10 | 2 | 61 | 3698 | 221914 |
| Nadeem | 2014-03-26 09:33:43 | 4 | 109 | 6565 | 393901 |
+-----------+---------------------+------+------+------+--------+
4 rows in set (0.00 sec)
DateInterval很好,但它有几个注意事项:
仅适用于PHP 5.3+(但这真的不再是一个好借口)仅支持年、月、日、小时、分钟和秒(无周)它计算上述所有+天的差异(你不能只计算月的差异)
为了克服这个问题,我编写了以下代码(由@enobrev答案改进而来):
function date_dif($since, $until, $keys = 'year|month|week|day|hour|minute|second')
{
$date = array_map('strtotime', array($since, $until));
if ((count($date = array_filter($date, 'is_int')) == 2) && (sort($date) === true))
{
$result = array_fill_keys(explode('|', $keys), 0);
foreach (preg_grep('~^(?:year|month)~i', $result) as $key => $value)
{
while ($date[1] >= strtotime(sprintf('+%u %s', $value + 1, $key), $date[0]))
{
++$value;
}
$date[0] = strtotime(sprintf('+%u %s', $result[$key] = $value, $key), $date[0]);
}
foreach (preg_grep('~^(?:year|month)~i', $result, PREG_GREP_INVERT) as $key => $value)
{
if (($value = intval(abs($date[0] - $date[1]) / strtotime(sprintf('%u %s', 1, $key), 0))) > 0)
{
$date[0] = strtotime(sprintf('+%u %s', $result[$key] = $value, $key), $date[0]);
}
}
return $result;
}
return false;
}
它运行两个循环;第一个算法通过暴力强制处理相对间隔(年和月),第二个算法通过简单的算法计算额外的绝对间隔(因此速度更快):
echo humanize(date_dif('2007-03-24', '2009-07-31', 'second')); // 74300400 seconds
echo humanize(date_dif('2007-03-24', '2009-07-31', 'minute|second')); // 1238400 minutes, 0 seconds
echo humanize(date_dif('2007-03-24', '2009-07-31', 'hour|minute|second')); // 20640 hours, 0 minutes, 0 seconds
echo humanize(date_dif('2007-03-24', '2009-07-31', 'year|day')); // 2 years, 129 days
echo humanize(date_dif('2007-03-24', '2009-07-31', 'year|week')); // 2 years, 18 weeks
echo humanize(date_dif('2007-03-24', '2009-07-31', 'year|week|day')); // 2 years, 18 weeks, 3 days
echo humanize(date_dif('2007-03-24', '2009-07-31')); // 2 years, 4 months, 1 week, 0 days, 0 hours, 0 minutes, 0 seconds
function humanize($array)
{
$result = array();
foreach ($array as $key => $value)
{
$result[$key] = $value . ' ' . $key;
if ($value != 1)
{
$result[$key] .= 's';
}
}
return implode(', ', $result);
}
简单的功能
function time_difference($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
我想带来一个稍微不同的视角,这似乎没有被提及。
你可以用声明的方式解决这个问题(就像任何其他问题一样)。重点是问你需要什么,而不是如何到达那里。
在这里,你需要与众不同。但这有什么不同?这是一个间隔,正如在最受欢迎的答案中所提到的。问题是如何获取它。您可以不显式调用diff()方法,而是按开始日期和结束日期创建一个间隔,即按日期范围:
$startDate = '2007-03-24';
$endDate = '2009-06-26';
$range = new FromRange(new ISO8601DateTime($startDate), new ISO8601DateTime($endDate));
所有诸如闰年之类的复杂问题都已经解决了。现在,当您有一个固定开始日期时间的间隔时,您可以获得一个人类可读的版本:
var_dump((new HumanReadable($range))->value());
它输出的正是你所需要的。
如果您需要一些自定义格式,这也不是问题。您可以使用ISO8601格式化类,该类接受具有六个参数的调用:年、月、日、小时、分钟和秒:
(new ISO8601Formatted(
new FromRange(
new ISO8601DateTime('2017-07-03T14:27:39+00:00'),
new ISO8601DateTime('2018-07-05T14:27:39.235487+00:00')
),
function (int $years, int $months, int $days, int $hours, int $minutes, int $seconds) {
return $years >= 1 ? 'More than a year' : 'Less than a year';
}
))
->value();
它的产量超过一年。
有关此方法的更多信息,请查看快速入门条目。