Question

如何确认并通过出生日期显示其年龄？

if ($_POST['DateOfBirth_Month'] == 'January') {
     $DateOfBirth_Month = $_POST['DateOfBirth_Month'];// January
} 
else if ($_POST['DateOfBirth_Month'] == 'February') {
     $DateOfBirth_Month = $_POST['DateOfBirth_Month'];// February
}

if ($_POST['DateOfBirth_Day'] == '1') {
    $DateOfBirth_Day = $_POST['DateOfBirth_Day'];// 1
} 
else if ($_POST['DateOfBirth_Day'] == '2') {
    $DateOfBirth_Day = $_POST['DateOfBirth_Day'];// 2
}



if(is_numeric($_POST['DateOfBirth_Year']) && $_POST['DateOfBirth_Year'] > 1969)
{     
    $DateOfBirth_Year = $_POST['DateOfBirth_Year'];
 } 
else if (is_numeric($_POST['DateOfBirth_Year']) && $_POST['DateOfBirth_Year'] < 1970)
{
     echo'<p><font color ="red">Please enter 1970 and above!</font></p>'; 
} 
else 
{
     echo '<p><font color="red">Enter NUMBER for the year of date of birth!</font></p>'; 
}

Answer 1

我个人更喜欢使用JS在客户端进行验证。您可以尝试使用正则表达式

/^\d{2}\/\d{2}\/\d{4}$/

验证日期。您可能会从this fiddle

中获得一些想法

Answer 2

我们假设我发布了我的生日：

$_POST["DateOfBirth_Year"] = 1980;
$_POST["DateOfBirth_Month"] = "October";
$_POST["DateOfBirth_Day"] = 20;

我使用此功能处理POST

function POST($var) {
  if(!isset($_POST[$var])) return null;
  return $_POST[$var];
}

首先我们阅读输入

$dob_m = POST('DateOfBirth_Month');
$dob_y = intval(POST('DateOfBirth_Year'));
// lets allow say 24th to be 24
$dob_d = intval(POST('DateOfBirth_Day'));

为了便于检查和获取月份索引，让我们创建月份数组

$months = array("January", "February", "March", "April", "May", "June", "July",
    "August", "September", "October", "November", "December");

现在检查月份

$month_ok = array_search($dob_m,$months)!==false;
if(!$month_ok) return; // some reasonable error handling should be here

...并获得月份指数

$dob_m = array_search($dob_m,$months) + 1;

然后我们应该使用date函数验证日期（30.2.1980无效）

$stamp = mktime(0,0,0,$dob_m,$dob_d,$dob_y);
$dob_ok = date("Y",$stamp)==$dob_y &&
    date("n",$stamp)==$dob_m && date("j",$stamp)==$dob_d;
if(!$dob_ok) return; // proper error handling here, too

最后，让我们使用DateTime对象

计算年龄

$dob_dt = new DateTime(date("Y-m-d H:i:s",$stamp));
$now_dt = new DateTime("now");
$age = $dob_dt->diff($now_dt)->format("%y");

if($age<0 || $age>125) return; // error handling here
echo $age; // should be my age

如何为下拉列表设置月和日的验证

2 个答案: