Heys guys,我一直在玩這個ajax電話幾天。我之前發佈了一個沒有變量的電子郵件,但現在我無法接到調用提交變量或不提供變量的形式。這裏是我的代碼:使用ajax發送信息至submit.php
<form id="contactform" method="post" action="submit.php">
<input type="text" placeholder="Name" name="username" id="username" />
<input type="email" placeholder="Your Email Address" name="email" id="email" />
<input type="button" id="submit" class="submit" value="submit"></a>
</form>
AJAX:
<script type="text/javascript">
$("#submit").click(function(event){
var data = $('#contactform').serialize();
event.preventDefault();
$.ajax({
url: "submit.php",
type: "POST",
data: data,
success: function() {
alert("Success!");
}
});
return false;
});
});
</script>
submit.php:
<?php
$from = "[email protected]";
$usersubject = "Thank You!";
$usermessage = "Thank you for signing up!";
$to = $_REQUEST['email'];
$subject = "Form Info";
?>
<?php
$name = $_REQUEST['username'];
$email = $_REQUEST['email'];
$message = "Name: $name
Email: $email";
$headers = "From:" . $from;
//mail($to,$subject,$message,$headers);
mail($email,$usersubject,$usermessage,$headers);
echo 'Success';
?>
當你直接用變量來訪問它(submit.php工作www.website.com /submit.php?username=Dave & [email protected])。在這裏有一個錯誤,我錯過了嗎?任何幫助表示讚賞!
這一個額外的'});',是不是因爲你錯過了包括'的document.ready '在這個問題中,或者這是如何在你的實現? – vee