Question

我有一个日期进入用户指定的循环。日期将始终来自格式化为“Y-m-d”字符串的数据库。我知道我可以直接比较字符串，只要它们是那种格式，但是，我也尝试使用strtotime转换日期来比较它们没有运气。我正在尝试确定用户在付款到期前有多少薪水

这就是我所拥有的

    $due_date = '2016-12-13';

    //count paychecks set to zero and added to by loop
    $paychecks = 0;

    //users next paycheck ('Y-m-d' ALWAYS)
    $next_payday = $user['next_payday']; //equal to '2016-12-02'

    //how often they get paid (int)
    $frequency = 14;

    while(strtotime($next_payday) <= strtotime($due_date)){

        //next_payday equals 1480654800 when coming into the loop
        //due_date equals 1481605200 when coming into the loop

        //add 14 days to the date
        $next_payday = date('Y-m-d', strtotime("+" .$frequency." days"));;

        //add to paychecks
        $paychecks++;
    }

问题是循环永远不会停止。它一直在前进和前进。

感谢任何人都能给我的帮助。

Answer 1

啊，一定要使用strtotime获取整数（表示epoch以来的秒数）进行比较，然后将天数乘以一天中的秒数（86400）：

$due_date = strtotime('2016-12-25');

//count paychecks set to zero and added to by loop
$paychecks = 0;

//users next paycheck (unixtime for comparison)
$next_payday = strtotime($user['next_payday']);

//how often they get paid (int)
$frequency = 14;

while($next_payday <= $due_date){

    //add 14 days to the date
    $next_payday += ($frequency * 86400);

    //add to paychecks
    $paychecks++;
}

在while循环中添加日期

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