我正在尝试转换格式为2009-09-12 20:57:19的时间戳，并将其转换为3分钟前用PHP。
我找到了一个有用的脚本来做这件事，但我认为它正在寻找一种不同的格式来用作时间变量。我想修改的脚本与此格式的工作是:
function _ago($tm,$rcs = 0) {
$cur_tm = time();
$dif = $cur_tm-$tm;
$pds = array('second','minute','hour','day','week','month','year','decade');
$lngh = array(1,60,3600,86400,604800,2630880,31570560,315705600);
for($v = sizeof($lngh)-1; ($v >= 0)&&(($no = $dif/$lngh[$v])<=1); $v--); if($v < 0) $v = 0; $_tm = $cur_tm-($dif%$lngh[$v]);
$no = floor($no);
if($no <> 1)
$pds[$v] .='s';
$x = sprintf("%d %s ",$no,$pds[$v]);
if(($rcs == 1)&&($v >= 1)&&(($cur_tm-$_tm) > 0))
$x .= time_ago($_tm);
return $x;
}
我认为在前几行脚本试图做的事情看起来像这样(不同的日期格式数学):
$dif = 1252809479 - 2009-09-12 20:57:19;
如何将我的时间戳转换成那种(unix?)格式?
一些语言显示时间之前有一些问题，例如阿拉伯语，有3种格式需要显示日期。
我在我的项目中使用这个函数，希望他们能帮助到别人(任何建议或改进我都会很感激:))
/**
*
* @param string $date1
* @param string $date2 the date that you want to compare with $date1
* @param int $level
* @param bool $absolute
*/
function app_date_diff( $date1, $date2, $level = 3, $absolute = false ) {
$date1 = date_create($date1);
$date2 = date_create($date2);
$diff = date_diff( $date1, $date2, $absolute );
$d = [
'invert' => $diff->invert
];
$diffs = [
'y' => $diff->y,
'm' => $diff->m,
'd' => $diff->d
];
$level_reached = 0;
foreach($diffs as $k=>$v) {
if($level_reached >= $level) {
break;
}
if($v > 0) {
$d[$k] = $v;
$level_reached++;
}
}
return $d;
}
/**
*
*/
function date_timestring( $periods, $format = 'latin', $separator = ',' ) {
$formats = [
'latin' => [
'y' => ['year','years'],
'm' => ['month','months'],
'd' => ['day','days']
],
'arabic' => [
'y' => ['سنة','سنتين','سنوات'],
'm' => ['شهر','شهرين','شهور'],
'd' => ['يوم','يومين','أيام']
]
];
$formats = $formats[$format];
$string = [];
foreach($periods as $period=>$value) {
if(!isset($formats[$period])) {
continue;
}
$string[$period] = $value.' ';
if($format == 'arabic') {
if($value == 2) {
$string[$period] = $formats[$period][1];
}elseif($value > 2 && $value <= 10) {
$string[$period] .= $formats[$period][2];
}else{
$string[$period] .= $formats[$period][0];
}
}elseif($format == 'latin') {
$string[$period] .= ($value > 1) ? $formats[$period][1] : $formats[$period][0];
}
}
return implode($separator, $string);
}
function timeago( $date ) {
$today = date('Y-m-d h:i:s');
$diff = app_date_diff($date,$today,2);
if($diff['invert'] == 1) {
return '';
}
unset($diff[0]);
$date_timestring = date_timestring($diff,'latin');
return 'About '.$date_timestring;
}
$date1 = date('Y-m-d');
$date2 = '2018-05-14';
$diff = timeago($date2);
echo $diff;
试试这个，我从我的旧代码中找到的，它显示了正确的结果
function ago($datefrom, $dateto = -1) {
// Defaults and assume if 0 is passed in that
// its an error rather than the epoch
if ($datefrom == 0) {
return "A long time ago";
}
if ($dateto == -1) {
$dateto = time();
}
// Make the entered date into Unix timestamp from MySQL datetime field
$datefrom = strtotime($datefrom);
// Calculate the difference in seconds betweeen
// the two timestamps
$difference = $dateto - $datefrom;
// Based on the interval, determine the
// number of units between the two dates
// From this point on, you would be hard
// pushed telling the difference between
// this function and DateDiff. If the $datediff
// returned is 1, be sure to return the singular
// of the unit, e.g. 'day' rather 'days'
switch (true) {
// If difference is less than 60 seconds,
// seconds is a good interval of choice
case(strtotime('-1 min', $dateto) < $datefrom):
$datediff = $difference;
$res = ($datediff == 1) ? $datediff . ' second' : $datediff . ' seconds';
break;
// If difference is between 60 seconds and
// 60 minutes, minutes is a good interval
case(strtotime('-1 hour', $dateto) < $datefrom):
$datediff = floor($difference / 60);
$res = ($datediff == 1) ? $datediff . ' minute' : $datediff . ' minutes';
break;
// If difference is between 1 hour and 24 hours
// hours is a good interval
case(strtotime('-1 day', $dateto) < $datefrom):
$datediff = floor($difference / 60 / 60);
$res = ($datediff == 1) ? $datediff . ' hour' : $datediff . ' hours';
break;
// If difference is between 1 day and 7 days
// days is a good interval
case(strtotime('-1 week', $dateto) < $datefrom):
$day_difference = 1;
while (strtotime('-' . $day_difference . ' day', $dateto) >= $datefrom) {
$day_difference++;
}
$datediff = $day_difference;
$res = ($datediff == 1) ? 'yesterday' : $datediff . ' days';
break;
// If difference is between 1 week and 30 days
// weeks is a good interval
case(strtotime('-1 month', $dateto) < $datefrom):
$week_difference = 1;
while (strtotime('-' . $week_difference . ' week', $dateto) >= $datefrom) {
$week_difference++;
}
$datediff = $week_difference;
$res = ($datediff == 1) ? 'last week' : $datediff . ' weeks';
break;
// If difference is between 30 days and 365 days
// months is a good interval, again, the same thing
// applies, if the 29th February happens to exist
// between your 2 dates, the function will return
// the 'incorrect' value for a day
case(strtotime('-1 year', $dateto) < $datefrom):
$months_difference = 1;
while (strtotime('-' . $months_difference . ' month', $dateto) >= $datefrom) {
$months_difference++;
}
$datediff = $months_difference;
$res = ($datediff == 1) ? $datediff . ' month' : $datediff . ' months';
break;
// If difference is greater than or equal to 365
// days, return year. This will be incorrect if
// for example, you call the function on the 28th April
// 2008 passing in 29th April 2007. It will return
// 1 year ago when in actual fact (yawn!) not quite
// a year has gone by
case(strtotime('-1 year', $dateto) >= $datefrom):
$year_difference = 1;
while (strtotime('-' . $year_difference . ' year', $dateto) >= $datefrom) {
$year_difference++;
}
$datediff = $year_difference;
$res = ($datediff == 1) ? $datediff . ' year' : $datediff . ' years';
break;
}
return $res;
}
示例:echo ago('2020-06-03 00:14:21 AM');
产出:6天
