檢查今天的日期是否在其他兩個日期之間

Question

我正在嘗試檢查今天的日期是否在某個時期的開始和停止日期之間，例如冬天，夏天，春天等。

並且，如果今天的日期介於冬季之間，則它將$ season變量設置為該日期。

但是目前它只給我“ 01/01”，我不明白為什么。

感謝幫助！ :)

$season = date("d-m");
$season = date("d-m", strtotime($season));


$startSummer = date("01-06");
$endSummer = date("31-08");

$startAutum = date("01-09");
$endAutum = date("30-11");

$startSpring = date("01-03");
$endSpring = date("31-05");

$startWinter = date("01-12");
$endWinter = date("28-02");

// start and stop, periods

// $startYear = date("d-m", strtotime($startYear));         $endYear = date("d-m", strtotime($endYear));
$startSummer = date("d-m", strtotime($startSummer));      $endSummer = date("d-m", strtotime($endSummer));
$startAutum = date("d-m", strtotime($startAutum));        $endAutum = date("d-m", strtotime($endAutum));
$startSpring = date("d-m", strtotime($startSpring));      $endSpring = date("d-m", strtotime($endSpring));
$startWinter = date("d-m", strtotime($startWinter));      $endWinter = date("d-m", strtotime($endWinter));

  if(($season > $startSummer) && ($season < $endSummer)){
    $season = "Sommar";
  }else if(($season > $startAutum) && ($season < $endAutum)){
    $season = "Höst";
  }else if(($season > $startSpring) && ($season < $endSpring)){
    $season = "Vår";
  }else if(($season > $startWinter) && ($season < $endWinter)){
    $season = "Vinter";
  }

Answer 1

您可以堅持使用時間戳。 不要轉換回日期。 您正在進行無效的比較，例如30-01小於28-02的假設。 計算機將比較第一個3和2，然后告訴您30-01正確大於28-02。 所以...

$startSummer = mktime(0,0,0, 6, 1, 2000); // The year doesn't matter according to your code
$endSummer = mktime(0,0,0, 8, 31, 2000);

現在，兩者之間有約會嗎？ 假設我正在檢查$ month和$ day ...

$myday = mktime(0,0,0, $month, $day, 2000);
if($myday>=$startSummer && $myday<=$endSummer) $season = "Summer";

Answer 2

如果使用DateTime對象（迄今為止最好的方法），則可以將它們與常規比較器進行比較，例如：

$date1 = new DateTime('today');
$date2 = new DateTime('2014-04-04');

if ($date1 < $date2) echo 'Past';
else if ($date1 == $date2) echo 'Present';
else echo 'Future';

參見文檔： http : //php.net/manual/en/datetime.diff.php#example-2368

Answer 3

請記住，變量可以被覆蓋-正如一年中的季節變化一樣，您的變量也可以被覆蓋-只要我們這樣做，我們就可以得出正確的變量。 這意味着我們僅需測試日期是否在季節更改的日期之后 。

// Since we're testing today's date
// we use the current year timestamps
$year = date('Y');
$startSpring = strtotime("$year-03-01");
$startSummer = strtotime("$year-06-01");
$startAutum = strtotime("$year-09-01");
$startWinter = strtotime("$year-12-01");

$today = time();

// The year starts with Winter
$season = 'Winter';
if($today > $startSpring) $season = 'Spring'; // Past the 1st day of spring?
if($today > $startSummer) $season = 'Summer'; // Etc...
if($today > $startAutumn) $season = 'Autumn';
if($today > $startWinter) $season = 'Winter';

echo 'It is currently '.$season;

這是在漂亮的函數中清除的相同邏輯，它將為您檢查任何日期並返回季節：

// Accepts an optional unix timestamp
// Uses the current date by default
function getSeason($test_date=FALSE){
    $test_date = $test_date ? $test_date : time();

    // Use the year of the date we're testing
    $year = date('Y', $test_date);

    // The year starts with Winter
    $season = 'Winter';
    if($test_date > strtotime("$year-03-01")) $season = 'Spring'; // Past the 1st day of spring?
    if($test_date > strtotime("$year-06-01")) $season = 'Summer'; // Etc...
    if($test_date > strtotime("$year-09-01")) $season = 'Autumn';
    if($test_date > strtotime("$year-12-01")) $season = 'Winter';

    return $season;
}