PHP Date and day difference function

Source: Internet
Author: User
Tags current time

Find the difference function of 22 date

<?php
$time 1 = "2008-6-15 11:49:59";//The first time
$time 2 = "2007-5-5 12:53:28";/The second time
$t 1 = strtotime ($time 1);
$t 2 = strtotime ($time 2);
$t = ABS ($t 1-$t 2);
$start = 0;
$string = "Two time difference:";
$y = Floor ($t 12/(3600*24*360));
if ($start | | $y)
{
$start = 1;
$t-= $y *3600*24*360;
$string. = $y. " Years ";
}
$m = Floor ($t 12/(3600*24*31));
if ($start | | $m)
{
$start = 1;
$t-= $m *3600*24*31;
$string. = $m. " Month ";
}
$d = Floor ($t 12/(3600*24));
if ($start | | $d)
{
$start = 1;
$t-= $d *3600*24;
$string. = $d. " Days ";
}
$h = Floor ($t 12/(3600));
if ($start | | $h)
{
$start = 1;
$t-= $h *3600;
$string. = $h. " When ";
}
$s = floor ($t 12/(60));
if ($start | | $s)
{
$start = 1;
$t-= $s *60;
$string. = $s. " Points ";
}
$string. = "{$t 12} seconds";
Echo $string;
?>

This is a function that asks for the difference of any time.

?
#作者: Xian le
#功能: To get a time difference between any and current time
function Querydays ($DATESTR) {
#格式化时间
$da =preg_split ("/(-| |:) /i ", $datestr);
$nowyear =date ("Y");
$nowmon =date ("n");
$nowday =date ("D");
$nowtimes =mktime (0,0,0, $nowmon, $nowday, $nowyear);
$pdtimes = Mktime (0,0,0, $nowmon, $nowday, $nowyear-1);
$bjtimes = Mktime (0,0,0, $da [1], $da [2], $da [0]);
#判断所给出的时间是不是在一年内
if ($bjtimes >= $pdtimes and $bjtimes <= $nowtimes) {
Return (Floor strftime ("%j", Mktime (0,0,0, $nowmon, $nowday, $nowyear)-mktime ($da [3], $da [4], $da [5], $da [1], $da [2],$ Da[0])));
}else{
$loop = $nowyear-$da [0];
$totaldays = (Floor (strftime ("%j", Mktime (0,0,0, $nowmon, $nowday, $nowyear)-mktime (0,0,0,1,1, $nowyear)));
for ($i =1; $i <= $loop; $i + +) {
For ($j =12 $j >=1; $j-) {
if ($da [0]== $nowyear-$i and $da [1]== $j) {
$days =mondays ($nowyear-$i, $j);
return $totaldays + + $days-$da [2];
Break
}else{
$days =mondays ($nowyear-$i, $j);
$totaldays + + $days;
}//end Else
}//end for
}//end for
}//end Else
}//end function
#取得月分的天数
Function Mondays ($year, $month) {
Switch ($month) {
Case "1":
Case "3":
Case "5":
Case "7":
Case "8":
Case "10":
Case "a": $days =31;break;
Case "4":
Case "6":
Case "9":
Case "One": $days =30;break;
Case "2":
if (Checkdate ($month, $year)) {
$days = 29;
}else{
$days = 28;
}//end Else
Break
}//end switch
return $days;
}//end function
$datestr = "2002-1-14 9:47:20";
echo querydays ($DATESTR);
?>

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