轮流值班逻辑

/*

每周一位，轮流值日。
获取某日轮到谁值日的逻辑如下：

1、获取某日属于本年的第几周，周的序号从1开始

2、设周序号为W

3、值班同事存为数组P，例如：['王建','马清秦','刘俊','覃光焰']

4、所有值班同事的人数即为数组P的长度，设这个长度为L

5、令 N = W % L, N = N === 0 ? L : N, 则有： 第N周值班的同事为P[N-1]

*/

// JavaScript实现

function getWeek (dateTime) {
        let temptTime = dateTime
        //周几
        let weekday = temptTime.getDay() || 7
        //周1+5天=周六
        temptTime.setDate(temptTime.getDate() - weekday + 1 + 5)
        let firstDay = new Date(temptTime.getFullYear(), 0, 1)
        let dayOfWeek = firstDay.getDay()
        let spendDay = 1
        if (dayOfWeek != 0) {
          spendDay = 7 - dayOfWeek + 1
        }
        firstDay = new Date(temptTime.getFullYear(), 0, 1 + spendDay)
        let d = Math.ceil((temptTime.valueOf() - firstDay.valueOf()) / 86400000)
        let result = Math.ceil(d / 7)
        return result + 1
}



function whoDuty(str, P){
  var time = new Date(str);
  var W = getWeek(time);
  var L = P.length;

  var N = W % L;

  N = N == 0 ? L : N;

  return P[N-1];
}

whoDuty("2020/2/28",['A','B','C','D'])

 

// PHP实现

function who_is_on_duty()
{
    $P = [
        '68649639@qq.com',
        '812154952@qq.com',
        '37534901@qq.com',
        '3141235825@qq.com'
    ];
    $W = date('W');
    $L = count($P);
    $N = $W % $L;
    $N = $N === 0 ? $L : $N;
    return $P[$N - 1];
}