PHP program to calculate the number of years, months and days between two times

PHP program to calculate the number of years, months and days between two times_PHP tutorial

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Release： 2016-07-13 10:44:05

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Our principle for calculating the time difference is based on the definition: 1. There are 360 days in a year and 30 days in a month; 2. 86400=24*60*60 in the code represents the number of seconds in a day, so it can be calculated

The effect is as follows

The code is as follows. What needs to be explained is: 1. Define a year as 360 days and a month as 30 days; 2. 86400=24*60*60 in the code represents the number of seconds in a day; 3. These two The time must be written in a standardized format similar to 2013-07-28; 4. Extended to all PHP programs, you can change the Get_option('swt_builddate') WordPress parameter for obtaining background data into the time parameter that needs to be compared.

代码如下	复制代码
//Get detail gap of year,month and days between two different time by vfhky 20130728 $common = (time()-strtotime(get_option('swt_builddate'))); $a = floor($common/86400/360); //整数年 $b = floor($common/86400/30) - $a12; //整数月 $c = floor($common/86400) - $a360 - $b*30; //整数日 $d = floor($common/86400); //总的天数 echo $a."年".$b."月".$c."日（共计".$d."天）"; ?>

Some other methods

The code is as follows

代码如下	复制代码
function count_days($a,$b){ $a_dt=getdate($a); $b_dt=getdate($b); $a_new=mktime(12,0,0,$a_dt['mon'],$a_dt['mday'],$a_dt['year']); $b_new=mktime(12,0,0,$b_dt['mon'],$b_dt['mday'],$b_dt['year']); return round(abs($a_new-$b_new)/86400); } //今天与2008年10月11日相差多少天 $date1=strtotime(time()); $date1=strtotime('10/11/2008'); $result=count_days($date1,$date2); echo $result; ?>

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代码如下	复制代码
//今天与2008年9月9日相差多少天 $Date_1=date("Y-m-d"); $Date_2="2008-10-11"; $d1=strtotime($Date_1); $d2=strtotime($Date_2); $Days=round(($d2-$d1)/3600/24); echo "今天与2008年10月11日相差".$Days."天"; ?>

function count_days($a,$b){
$a_dt=getdate($a);
$b_dt=getdate($b);
$a_new=mktime(12,0,0,$a_dt['mon'],$a_dt['mday'],$a_dt['year']);
$b_new=mktime(12,0,0,$b_dt['mon'],$b_dt['mday'],$b_dt['year']);
return round(abs($a_new-$b_new)/86400);
}
//How many days are there between today and October 11, 2008
$date1=strtotime(time());
$date1=strtotime('10/11/2008');
$result=count_days($date1,$date2);
echo $result;
?>

Example 2

The code is as follows

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//How many days are there between today and September 9, 2008
$Date_1=date("Y-m-d");
$Date_2="2008-10-11";
$d1=strtotime($Date_1);
$d2=strtotime($Date_2);
$Days=round(($d2-$d1)/3600/24);
echo "The difference between today and October 11, 2008".$Days."Days";
?>

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www.bkjia.comtruehttp: //www.bkjia.com/PHPjc/633125.htmlTechArticleOur principle for calculating the time difference is based on the definition 1. A year is 360 days and a month is 30 days; 2. Code 86400=24*60*60, represents how many seconds there are in a day, so you can calculate the effect as...