Question

任何人都可以看到我的PHP表单发件人代码有什么问题：??

我的代码列在下面 - 表单总是说不管是什么，不正确的电子邮件地址。因此，它被打破了。非常感谢你的帮助！

$to = "joinrendezvous@gmail.com";
$from = "mail@example.com";

$headers = "From: " . $from . "\r\n";

$subject = "New subscription";
$body = "New user subscription: " . $_POST['email'];


if (filter_var($_POST['email'], FILTER_VALIDATE_EMAIL)) { 

    if (mail($to, $subject, $body, $headers, "-f " . $from)) {
        echo 'Alright ! You will be notified on <b>  ' . $_POST['email'] . '</b> :)';
    }
    else {
        echo 'There was a problem with your e-mail (' . $_POST['email'] . ')';  
    }
}
else {
    echo 'There was a problem with your e-mail (' . $_POST['email'] . ')';  
}

Answer 1

什么是错误的电子邮件＆amp;你想做什么？

看看这个：

if (filter_var($_POST['email'], FILTER_VALIDATE_EMAIL)) { 

    if (mail($to, $subject, $body, $headers, "-f " . $from)) {

if检查$_POST['email']时，mail命令正在使用$to作为变量。咦？也许你的代码应该是：

if (filter_var($to, FILTER_VALIDATE_EMAIL)) { 

    if (mail($to, $subject, $body, $headers, "-f " . $from)) {
        echo 'Alright ! You will be notified on <b>  ' . $to . '</b> :)';
    }
    else {
        echo 'There was a problem with your e-mail (' . $to . ')';  
    }
}
else {
    echo 'There was a problem with your e-mail (' . $to . ')';  
}

但是，如果您不打算使用$_POST表单变量，为什么还要通过表单执行此操作？

Answer 2

我认为问题是$ _POST ['email']。回复它，这样你就可以看到$ _POST ['email']的电子邮件格式是否正确，因为filter_var检查了..

如何让这个工作PHP表单发送者

2 个答案: