从日期字符串中获取年份

Question

我想检查当前年份是否大于日期字符串（DMY），这是我的代码

$OldDate = "09-30-2011";
$OldYear = strtok($OldDate, '-');
$NewYear = date("Y");

if ($OldYear < $NewYear) {
    echo "Year is less than current year"   
} else {
    echo "Year is greater than current year";
}

Answer 1

您可以使用strtotime()：

$OldDate = "2011-09-30";

$oldDateUnix = strtotime($OldDate);
if(date("Y", $oldDateUnix) < date("Y")) {
    echo "Year is less than current year";
} else {
    echo "Year is greater than current year";
}

更新

因为您使用的是非常规日期戳，所以您必须使用不同的方法，例如：

$OldDate = "09-30-2011";
list($month, $day, $year) = explode("-", $OldDate);
$oldDateUnix = strtotime($year . "-" . $month . "-" . $day);
if(date("Y", $oldDateUnix) < date("Y")) {
    echo "Year is less than current year";
} else {
    echo "Year is greater than current year";
}

注意：如果您想始终确保您的日期被 正确理解strtotime，请使用YYYY-MM-DD