PHP获取时间排除周六、周日的两个方法_php实例

//方法一：
<?php
$now = time(); //指定日期用法 $now = strtotime('2014-01-08') ;
$day = 3600*24;
$total = 12;

$days =array() ;

for ($i=2;$i<$total;$i++)
{
    $timer = $now+$day*$i;
    $num= date("N",$timer)-2; //周一开始
    if($num>=-1 and $num<=3)
    {
        if(count($days)>=10) break;
        $days[]=date("Y-m-d",$now+$day*$i);
        $total +=1 ;// $total==12 ?$total+1:$total;

    }else
    {
        $total = $total==12 ?$total+1:$total;
    }
}
$i=1;
foreach($days as $day)
{

    echo "$i===>".$day."\n";
    $i++;
}


//方法二：
function get_days ($date="")
{
    $now = empty($date)?time():strtotime($date);
    $days = array();
    $i = 2;
    while(count($days)<10)
    {
        $timer = $now+3600*24*$i;
        $num= date("N",$timer)-2; //周一开始
        if($num>=-1 and $num<=3)
        {
            $days[]=date("Y-m-d",$now+3600*24*$i);
        }
        $i++;
    }

 return $days;
}