php - 检查是否超过 18 年，仅适用于 1960->1994

Question

我的表格有一些问题。 PHP 将检查年龄是否未满 18 岁，如果是，则此人会收到一条消息，提示他需要年满 18 岁。但是，该表单仅在此人出生于 1960 年至 1994 年之间时才有效。如果此人出生于 1950 年，他会收到一条消息说他未满 18 岁...

我的代码有什么问题？

if(empty($_POST["born"])) {$born_error = ('<span class="notutfyllt">*Fødselsår is not filled out</span><br/>');}
else {
        $today = date_create('today');
        $born = date_create("{$_POST['born']}");
        $age = date_diff($today, $born)->y;

        } if ($age >= 18) {
            $born_error = "";
        } else { 
            $born_error = ('<span class="notutfyllt">*You need to be 18 year</span><br/>');
        }

来自表单:

<label for="born">Year of born:</label>
      <input type="text" name="born" id="born" value="'.$born.'">

新版本:

if(empty($_POST["born"])) {$born_error = ('<span class="notutfyllt">*Fødselsår is not filled out</span><br/>');}
    else {
        $bday = new DateTime($_POST["born"]);
        $today = new DateTime('Y');
        $diff = $today->diff($bday);
        if($diff->y>18) {
            $born_error = "";
        }else {
            $born_error = ('<span class="notutfyllt">*You need to be 18 years</span><br/>');
        }
    }

Answer 1

来自 PHP manual 的注释:

The "Year (and just the year)" format only works if a time string has already been found -- otherwise this format is recognised as HH MM.

例子:

$born = date_create('1950');
var_dump($born);

产生:

object(DateTime)#1 (3) {
  ["date"]=>
  string(19) "2013-12-05 19:50:00"
  ["timezone_type"]=>
  int(3)
  ["timezone"]=>
  string(13) "Europe/Berlin"
}

我猜1960及以上是无效时间，所以算作年。您将不得不询问完整的出生日期或在您的脚本中添加 "-01-01"(这会给某些人带来错误的结果)。