本文實例講述了php中strtotime函數用法。分享給大家供大家參考。具體如下：
strtotime(字符串$時間[，詮釋$現在])int strtotime(string $time [,int $now] 該函數期望得到一個包含美國英語日期格式,并會嘗試解析成一個Unix時間戳(多少秒自1970年1月1日00:00:00星期一該格式),相對于現在提供的時間戳,或當前時間如果現在不提供
這個函數將使用TZ環境變量(如果有)來計算時間戳,自PHP 5.1.0有更容易的方法來確定所使用的所有/日期時間函數的時區,這一過程是解釋在date_default_timezone_get()函數的一頁.
解析的字符串,在PHP 5.0.0，不允許在微秒的時間,自PHP 5.0.0他們是允許的,但忽略.
現在哪些是作為計算基數相對日期使用時間戳.
返回值:在成功返回一個時間戳,否則返回FALSE,前到PHP 5.1.0，這個函數將返回失敗-1.
現在我們來看看strtotime字符轉換成時間的函數實例,代碼如下:
復制代碼代碼如下:
<?php
//function
function nextWeeksDay($date_begin,$nbrweek)
{
$nextweek=array();
for($i = 1; $i <= $nbrweek; $i++) { // 52 week in one year of coursewww.phpfensi.com
$nextweek[$i]=date('D d M Y', strtotime('+'.$i.' week',$date_begin));
}
return $nextweek;
}
/// end function
/// example of a select date
// var
$date_begin = strtotime('06-05-2010'); //D Day Month Year - like function format.
$nbrweek=52;
// call function
$result=nextWeeksDay($date_begin,$nbrweek);
// Preview
for($i = 1; $i <= $nbrweek; $i++) {
echo '<br> - '.$result[$i];
}
?>
<?php
$str = 'Not Good';
// previous to PHP 5.1.0 you would compare with -1, instead of false
if (($timestamp = strtotime($str)) === false) {
echo "The string ($str) is bogus";
} else {
echo "$str == " . date('l dS o F Y h:i:s A', $timestamp);
}
?>
<?php
echo strtotime("now"), " ";
echo strtotime("10 September 2000"), " ";
echo strtotime("+1 day"), " ";
echo strtotime("+1 week"), " ";
echo strtotime("+1 week 2 days 4 hours 2 seconds"), " ";
echo strtotime("next Thursday"), " ";
echo strtotime("last Monday"), " ";
?>
//function
function nextWeeksDay($date_begin,$nbrweek)
{
$nextweek=array();
for($i = 1; $i <= $nbrweek; $i++) { // 52 week in one year of coursewww.phpfensi.com
$nextweek[$i]=date('D d M Y', strtotime('+'.$i.' week',$date_begin));
}
return $nextweek;
}
/// end function
/// example of a select date
// var
$date_begin = strtotime('06-05-2010'); //D Day Month Year - like function format.
$nbrweek=52;
// call function
$result=nextWeeksDay($date_begin,$nbrweek);
// Preview
for($i = 1; $i <= $nbrweek; $i++) {
echo '<br> - '.$result[$i];
}
?>
<?php
$str = 'Not Good';
// previous to PHP 5.1.0 you would compare with -1, instead of false
if (($timestamp = strtotime($str)) === false) {
echo "The string ($str) is bogus";
} else {
echo "$str == " . date('l dS o F Y h:i:s A', $timestamp);
}
?>
<?php
echo strtotime("now"), " ";
echo strtotime("10 September 2000"), " ";
echo strtotime("+1 day"), " ";
echo strtotime("+1 week"), " ";
echo strtotime("+1 week 2 days 4 hours 2 seconds"), " ";
echo strtotime("next Thursday"), " ";
echo strtotime("last Monday"), " ";
?>
這是一個快速函數計算在一年期間,“工作天”,“工作日”是指那些沒有周末,沒有假期在$數組中指定的假日,實例代碼如下:
復制代碼代碼如下:
function get_working_days($to_date) {
$holidays = array(
1 => array(10), //2011 ...
2 => array(11),
3 => array(21), //... 2011
4 => array(29,30), //2010 ...
5 => array(3,4,5),
6 => array(),
7 => array(19),
8 => array(11,12,13),
9 => array(20,23),
10 => array(11),
11 => array(3,23),
12 => array(23) //... 2010
);
for($to_date, $w = 0, $i = 0, $x = time(); $x < $to_date; $i++, $x = strtotime("+$i day")) {
if(date("N",$x) < 6 &! in_array(date("j",$x),$holidays[date("n",$x)])) $w++;
}
return $w;
}
//Usage:
echo get_working_days(strtotime("2011-01-08"));
$holidays = array(
1 => array(10), //2011 ...
2 => array(11),
3 => array(21), //... 2011
4 => array(29,30), //2010 ...
5 => array(3,4,5),
6 => array(),
7 => array(19),
8 => array(11,12,13),
9 => array(20,23),
10 => array(11),
11 => array(3,23),
12 => array(23) //... 2010
);
for($to_date, $w = 0, $i = 0, $x = time(); $x < $to_date; $i++, $x = strtotime("+$i day")) {
if(date("N",$x) < 6 &! in_array(date("j",$x),$holidays[date("n",$x)])) $w++;
}
return $w;
}
//Usage:
echo get_working_days(strtotime("2011-01-08"));
希望本文所述對大家的PHP程序設計有所幫助。
