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):
$dif = 1252809479 - 2009-09-12 20:57:19;
如何将我的时间戳转换成那种(unix?)格式?
这是我的解决方案，请检查并根据您的要求修改
function getHowLongAgo($date, $display = array('Year', 'Month', 'Day', 'Hour', 'Minute', 'Second'), $ago = '') {
date_default_timezone_set('Australia/Sydney');
$timestamp = strtotime($date);
$timestamp = (int) $timestamp;
$current_time = time();
$diff = $current_time - $timestamp;
//intervals in seconds
$intervals = array(
'year' => 31556926, 'month' => 2629744, 'week' => 604800, 'day' => 86400, 'hour' => 3600, 'minute' => 60
);
//now we just find the difference
if ($diff == 0) {
return ' Just now ';
}
if ($diff < 60) {
return $diff == 1 ? $diff . ' second ago ' : $diff . ' seconds ago ';
}
if ($diff >= 60 && $diff < $intervals['hour']) {
$diff = floor($diff / $intervals['minute']);
return $diff == 1 ? $diff . ' minute ago ' : $diff . ' minutes ago ';
}
if ($diff >= $intervals['hour'] && $diff < $intervals['day']) {
$diff = floor($diff / $intervals['hour']);
return $diff == 1 ? $diff . ' hour ago ' : $diff . ' hours ago ';
}
if ($diff >= $intervals['day'] && $diff < $intervals['week']) {
$diff = floor($diff / $intervals['day']);
return $diff == 1 ? $diff . ' day ago ' : $diff . ' days ago ';
}
if ($diff >= $intervals['week'] && $diff < $intervals['month']) {
$diff = floor($diff / $intervals['week']);
return $diff == 1 ? $diff . ' week ago ' : $diff . ' weeks ago ';
}
if ($diff >= $intervals['month'] && $diff < $intervals['year']) {
$diff = floor($diff / $intervals['month']);
return $diff == 1 ? $diff . ' month ago ' : $diff . ' months ago ';
}
if ($diff >= $intervals['year']) {
$diff = floor($diff / $intervals['year']);
return $diff == 1 ? $diff . ' year ago ' : $diff . ' years ago ';
}
}
谢谢
我知道这里有几个答案，但这是我想到的。这只处理MySQL DATETIME值，根据我回应的原始问题。数组$a需要做一些工作。我欢迎就如何改进提出意见。电话为:
Echo time_elapsed_string('2014-11-14 09:42:28');
function time_elapsed_string($ptime)
{
// Past time as MySQL DATETIME value
$ptime = strtotime($ptime);
// Current time as MySQL DATETIME value
$csqltime = date('Y-m-d H:i:s');
// Current time as Unix timestamp
$ctime = strtotime($csqltime);
// Elapsed time
$etime = $ctime - $ptime;
// If no elapsed time, return 0
if ($etime < 1){
return '0 seconds';
}
$a = array( 365 * 24 * 60 * 60 => 'year',
30 * 24 * 60 * 60 => 'month',
24 * 60 * 60 => 'day',
60 * 60 => 'hour',
60 => 'minute',
1 => 'second'
);
$a_plural = array( 'year' => 'years',
'month' => 'months',
'day' => 'days',
'hour' => 'hours',
'minute' => 'minutes',
'second' => 'seconds'
);
foreach ($a as $secs => $str){
// Divide elapsed time by seconds
$d = $etime / $secs;
if ($d >= 1){
// Round to the next lowest integer
$r = floor($d);
// Calculate time to remove from elapsed time
$rtime = $r * $secs;
// Recalculate and store elapsed time for next loop
if(($etime - $rtime) < 0){
$etime -= ($r - 1) * $secs;
}
else{
$etime -= $rtime;
}
// Create string to return
$estring = $estring . $r . ' ' . ($r > 1 ? $a_plural[$str] : $str) . ' ';
}
}
return $estring . ' ago';
}
