在 PHP 中检查闰年条件不起作用

Question

我试图弄清楚如何应用条件来检查一年是否是闰年。但是当我添加跳跃功能时，它就不起作用了。知道如何让它发挥作用吗？这是我的代码：    <?php$name = "";$character = "";$email = "";$birth_year = 1969;$validation_error = "";$existing_users = ["admin", "guest"];$options = ["options" => ["min_range" => 1920, "max_range" => date("Y")]];function leap($year) {    date('L', strtotime("$year-01-01")) ? TRUE : FALSE;}if ($_SERVER["REQUEST_METHOD"] == "POST") {  $raw_name = trim(htmlspecialchars($_POST["name"]));  if (in_Array($raw_name,$existing_users)){    $validation_error .= "This name is taken. <br>";  } else {      $name = $raw_name;    }  $raw_character = $_POST["character"];    if (in_array($raw_character,["wizard", "mage", "orc"])) {        $character = $raw_character;} else {    $validation_error .= "You must pick a wizard, mage, or orc. <br>";  }  $raw_email = $_POST["email"];  if (filter_var($raw_email,FILTER_VALIDATE_EMAIL)) {    $email = $raw_email;  } else {    $validation_error .= "Invalid email. <br>";  }  $raw_birth_year = $_POST["birth_year"];  if (filter_var($raw_birth_year,FILTER_VALIDATE_INT,$options)){$birth_year = $raw_birth_year;if ($raw_character === "mage") {    if (!leap($birth_year)){    $validation_error .= "Mages have to be born on leap years. <br>";  }}} else {  $validation_error .= "That can't be your birth year. <br>";}}?><h1>Create your profile</h1><form method="post" action=""><p>Select a name: <input type="text" name="name" value="<?php echo $name;?>" ></p><p>Select a character:  <input type="radio" name="character" value="wizard" <?php echo ($character=='wizard')?'checked':'' ?>> Wizard  <input type="radio" name="character" value="mage" <?php echo ($character=='mage')?'checked':'' ?>> Mage  <input type="radio" name="character" value="orc" <?php echo ($character=='orc')?'checked':'' ?>> Orc</p><p>Enter an email:<input type="text" name="email" value="<?php echo $email;?>" ></p><p>

Answer 1

我认为你的问题是该函数没有返回任何内容，请尝试：

function leap($year) { 
   return date('L', strtotime("$year-01-01")) ? TRUE : FALSE;
}