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 $val;
}
我知道这里有几个答案，但这是我想到的。这只处理MySQL DATETIME值，根据我回应的原始问题。数组$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';
}
