PHP没有迭代

Question

我有一个给定元素的数组：

date_default_timezone_set("UTC");
$DateNow = date("d/m/Y H:i", time());
$tommorowUnix = strtotime("+1 day");
$tommorowAfterUnix = strtotime("+2 day");
$holidays = array("30 October 2021",
"31 October 2021",
"01 November 2021",
"02 November 2021",
"11 November 2021",
"12 October 2021",
"25 December 2021",
"26 December 2021",
"27 December 2021",
"1 January 2022",
"2 January 2022");

然后，我有一个函数，将给定的$date 与数组进行比较，如果发现为真则返回：

foreach($holidays as &$value)
{
    $value = strtotime($value);
}

function isNextDayWeekend($date)
{
    $weekDay = date('w', $date);
    echo($weekDay);
    if($weekDay == 0 || $weekDay == 6)
        return true;
    else
        return false;   
}

function isNextDayHoliday($date)
{
    $returnVal = false;
    foreach ($holidays as $holidayDay)
    {
        echo ("test");
        if($date == $holidayDay) { $returnVal = true; }     
    }
    return $returnVal;
}
$check1 = isNextDayHoliday(strtotime("12 October 2021"));
echo $check1 ? 'true' : 'false';

很遗憾，连echo ("test") 都没有显示出来。

@EDIT：解决了上述情况。 $holidays 范围是个问题。尽管如此，它仍然给了我不好的价值观：

function isNextDayHoliday($date,$holidays)
{
    $returnVal = false;
    foreach ($holidays as $holidayDay)
    {
        if($date == $holidayDay) {$returnval = true;}   
        echo "Checking ".$date." vs. ".$holidayDay." = ";
        echo $returnVal ? "true" : "false"."<br/>";
    }
    return $returnVal;
}
$check1 = isNextDayHoliday(strtotime("12 October 2021"),$holidays);
echo $check1 ? "true" : "false";

Checking 1633996800 vs. 1635552000 = false
Checking 1633996800 vs. 1635638400 = false
Checking 1633996800 vs. 1635724800 = false
Checking 1633996800 vs. 1635811200 = false
Checking 1633996800 vs. 1636588800 = false
**Checking 1633996800 vs. 1633996800 = false**
Checking 1633996800 vs. 1640390400 = false
Checking 1633996800 vs. 1640476800 = false
Checking 1633996800 vs. 1640563200 = false
Checking 1633996800 vs. 1640995200 = false
Checking 1633996800 vs. 1641081600 = false
false

Answer 1

function isNextDayHoliday($date)
{
    $returnVal = false;
    foreach ($holidays as $holidayDay)
    {
        echo ("test");
        if($date == $holidayDay) { $returnVal = true; }     
    }
    return $returnVal;
}

你需要将$holidays传递给这个函数，这个函数不知道holidays是什么。

你可以全局声明，然后就可以使用了。

你来优化一下。 if($date == $holidayDay) return true; 以便在您找到它后立即停止处理。