结束日期验证

Question

我想验证当前日期与问题张贴日期，如果日期差异大于1天，那么它需要显示在过期的问题，这将不会考虑我的数组中已经有的日期数组。

比如一个问题发布在7月20日-11日。

豁免日期数组{‘21-7-11’，...etc}当前日期为22-7-11

然后，输出需要显示为问题等待1天而不是2天。

有谁能帮上忙吗？

Answer 1

这是我为计算到期日而创建的一些示例代码。提交问题时，以这种方式计算到期日，并将其与该问题的条目一起存储在数据库中...

<?php

$daysDue = 2;
$exDatesArray = array("2011-07-21", "2011-07-23"); //array with all your holiday dates.
$question_1_Date = "2011-07-21";
$question_2_Date = "2011-07-18";
$question_3_Date = "2011-07-20";

echo "Holidays on: ";
foreach( $exDatesArray AS $exdate )
{
    echo $exdate . ", ";
}
echo "<br/><br/>";
echo "submit date -> due date<br/>";
echo $question_1_Date . " -> " . calculateDueDate( $question_1_Date, $exDatesArray, $daysDue ) . "<br/>";
echo $question_2_Date . " -> " . calculateDueDate( $question_2_Date, $exDatesArray, $daysDue ) . "<br/>";
echo $question_3_Date . " -> " . calculateDueDate( $question_3_Date, $exDatesArray, $daysDue );

function calculateDueDate( $date, $exDates, $daysDue )
{
    //start with day 1
    $count = 1;

    //now we loop from day one to due days
    while( $count <= $daysDue )
    {
        //add one day to start date
        $date = add_date( $date, 1 );
        //check if that new date is a holiday
        if( !in_array($date , $exDates) )
        {
            //only if it is not a holiday we increase counter, otherwise we dont count that day towards due period
            $count++;
        }
    }

    //return calculated due date
    return $date;
}

function add_date($date,$days)
{ 
    $cd = strtotime($date); 
    return date('Y-m-d', mktime(0,0,0,date('m',$cd),date('d',$cd) + $days, date('Y',$cd))); 
}
?>

输出：

Holidays on: 2011-07-21, 2011-07-23, 

submit date -> due date
2011-07-21 -> 2011-07-24
2011-07-18 -> 2011-07-20
2011-07-20 -> 2011-07-24

Answer 2

//convert date to unixtime (seconds)
$question_sec = strtotime($question);

//subtract
$diff = time() - $question_sec;
$days = $diff / 60*60*24;
echo "$days old";