php判断某年某月有多少天

<?php
$year = 2001; //这里输入的年份
$month = 10;    //这里输入的月份

switch ($month) {
    case 1:
        echo "{$year}年{$month}月的天数是31天";
        break;
    case '3':
        echo "{$year}年{$month}月的天数是31天";
        break;
    case '5':
        echo "{$year}年{$month}月的天数是31天";
        break;
    case '7':
        echo "{$year}年,{$month}月的天数是31天";
        break;
    case '8':
        echo "{$year}年,{$month}月的天数是31天";
        break;
    case '10':
        echo "{$year}年,{$month}月的天数是31天";
        break;
    case '12':
        echo "{$year}年,{$month}月的天数是31天";
        break;

    case '4':
        echo "{$year}年,{$month}月的天数是30天";
        break;
    case '6':
        echo "{$year}年,{$month}月的天数是30天";
        break;
    case '9':
        echo "{$year}年,{$month}月的天数是30天";
        break;
    case '11':
        echo "{$year}年,{$month}月的天数是30天";
        break;
    case '2':
        if($year % 4 == 0 && $year % 100 != 0 || $year % 400 == 0 ){
            echo "{$year}年,{$month}月的天数是29天";
        }else{
            echo "{$year}年,{$month}月的天数是28天";
        }
        break;
    default:
        echo '你TM乱输入什么玩意,艹!';
        break;
}

?>





这里再交给大家一个简单的写法,就是天数一样的月份 让他break取消;直接穿越过去!

<?php
$year = 2001; //这里输入的年份
$month = 10;    //这里输入的月份

switch ($month) {
    case '1':
    case '3':
    case '5':
    case '7':
    case '8':
    case '10':
    case '12':
        echo "{$year}年,{$month}月的天数是31天";
        break;

    case '4':
    case '6':
    case '9':
    case '11':
        echo "{$year}年,{$month}月的天数是30天";
        break;

    case '2':
        if($year % 4 == 0 && $year % 100 != 0 || $year % 400 == 0 ){
            echo "{$year}年,{$month}月的天数是29天";
        }else{
            echo "{$year}年,{$month}月的天数是28天";
        }
        break;
        
    default:
        echo '你TM乱输入什么玩意,艹!';
        break;
}

?>

第二种看着是不是简单很多呢?!  嗯哼?