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):
$dif = 1252809479 - 2009-09-12 20:57:19;
如何将我的时间戳转换成那种(unix?)格式?
此函数不是为英语语言而设计的。我把这些单词翻译成英语。在用于英语之前，这需要更多的修正。
function ago($d) {
$ts = time() - strtotime(str_replace("-","/",$d));
if($ts>315360000) $val = round($ts/31536000,0).' year';
else if($ts>94608000) $val = round($ts/31536000,0).' years';
else if($ts>63072000) $val = ' two years';
else if($ts>31536000) $val = ' a year';
else if($ts>24192000) $val = round($ts/2419200,0).' month';
else if($ts>7257600) $val = round($ts/2419200,0).' months';
else if($ts>4838400) $val = ' two months';
else if($ts>2419200) $val = ' a month';
else if($ts>6048000) $val = round($ts/604800,0).' week';
else if($ts>1814400) $val = round($ts/604800,0).' weeks';
else if($ts>1209600) $val = ' two weeks';
else if($ts>604800) $val = ' a week';
else if($ts>864000) $val = round($ts/86400,0).' day';
else if($ts>259200) $val = round($ts/86400,0).' days';
else if($ts>172800) $val = ' two days';
else if($ts>86400) $val = ' a day';
else if($ts>36000) $val = round($ts/3600,0).' year';
else if($ts>10800) $val = round($ts/3600,0).' years';
else if($ts>7200) $val = ' two years';
else if($ts>3600) $val = ' a year';
else if($ts>600) $val = round($ts/60,0).' minute';
else if($ts>180) $val = round($ts/60,0).' minutes';
else if($ts>120) $val = ' two minutes';
else if($ts>60) $val = ' a minute';
else if($ts>10) $val = round($ts,0).' second';
else if($ts>2) $val = round($ts,0).' seconds';
else if($ts>1) $val = ' two seconds';
else $val = $ts.' a second';
return $val;
}
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 $diff;
我想有一个荷兰版本，支持单复数。仅仅在结尾加一个“s”是不够的，我们用的是完全不同的词，所以我重写了这篇文章的顶部答案。
这将导致:
2年1个月2周1天1分2秒
or
1年2个月1周2天1分1秒
public function getTimeAgo($full = false){
$now = new \DateTime;
$ago = new \DateTime($this->datetime());
$diff = $now->diff($ago);
$diff->w = floor($diff->d / 7);
$diff->d -= $diff->w * 7;
$string = array(
'y' => 'jaren',
'm' => 'maanden',
'w' => 'weken',
'd' => 'dagen',
'h' => 'uren',
'i' => 'minuten',
's' => 'seconden',
);
$singleString = array(
'y' => 'jaar',
'm' => 'maand',
'w' => 'week',
'd' => 'dag',
'h' => 'uur',
'i' => 'minuut',
's' => 'seconde',
);
// M.O. 2022-02-11 I rewrote this function to support dutch singles and plurals. Added some docs for next programmer to break his brain :)
// For each possible notation, if corresponding value of current key is true (>1) otherwise remove its key/value from array
// If the value from current key is 1, use value from $singleString array. Otherwise use value from $string array
foreach ($string as $k => &$v) {
if ($diff->$k) {
if($diff->$k == 1){
$v = $diff->$k . ' ' . $singleString[$k];
} else {
$v = $diff->$k . ' ' . $v;
}
} else {
if($diff->$k == 1){
unset($singleString[$k]);
} else {
unset($string[$k]);
}
}
}
// If $full = true, print all values.
// Values have already been filtered with foreach removing keys that contain a 0 as value
if (!$full) $string = array_slice($string, 0, 1);
return $string ? implode(', ', $string) . '' : 'zojuist';
}
你应该先测试一下，因为我不是一个优秀的程序员:)
