我的php运行但由于某种原因我的变量没有被传达。我做错了什么？我试图通过ajax中继消息，我似乎无法得到任何类型的错误或成功消息弹出，无论我把它放在我的php ...这导致我相信问题在于我的内心ajax / javascript函数。 ajax应该将消息直接放在定义的内容中。我也意识到这已经被问到了这里，但我真的看了很多，但仍然无法弄清楚出了什么问题。谢谢你们，对不起墙。
AJAX
<!-- Email -->
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
<script>
// magic.js
$(document).ready(function() {
// process the form
$('form').submit(function(event) {
$('#sub_result').html("");
// get the form data
// there are many ways to get this data using jQuery (you can use the class or id also)
var formData = {
'email' : $('input[name=email]').val(),
};
// process the form
$.ajax({
type : 'POST', // define the type of HTTP verb we want to use (POST for our form)
url : 'phEmail.php', // the url where we want to POST
data : formData, // our data object
dataType : 'json', // what type of data do we expect back from the server
encode : true
})
// using the done promise callback
.done(function(data) {
// log data to the console so we can see
console.log(data);
// here we will handle errors and validation messages
if ( ! data.success) {
// handle errors for email ---------------
if (data.errors.email) {
$('#sub_result').addClass('class="error"'); // add the error class to show red input
$('#sub_result').append('<div class="error">' + data.errors.email + '</div>'); // add the actual error message under our input
}
} else {
// ALL GOOD! just show the success message!
$('#sub_result').append('<div class="success" >' + data.message + '</div>');
// usually after form submission, you'll want to redirect
// window.location = '/thank-you'; // redirect a user to another page
}
})
// using the fail promise callback
.fail(function(data) {
// show any errors
// best to remove for production
console.log(data);
});
// stop the form from submitting the normal way and refreshing the page
event.preventDefault();
});
});
</script>
PHP
<?php
$errors = array(); // array to hold validation errors
$data = array(); // array to pass back data
// validate the variables ======================================================
// if any of these variables don't exist, add an error to our $errors array
if(filter_var($_POST['email'],FILTER_VALIDATE_EMAIL) === false)
{
$errors['email'] = 'Email is not valid';
}
if (empty($_POST['email'])){
$errors['email'] = 'Email is required.';
}
// if there are items in our errors array, return those errors============================
if ( ! empty($errors)) {
$data['success'] = false;
$data['errors'] = $errors;
} else {
//variables===============================================================================
$servername = "localhost";
$username = "ghostx19";
$password = "nick1218";
$dbname = "ghostx19_samplepacks";
$user = $_POST['email'];
// Create connection======================================================================
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
echo "Connection failed";
}
//add user================================================================================
$sql = "INSERT INTO users (email)
VALUES ('$user')";
if ($conn->query($sql) === TRUE) {
$data['success'] = true;
$data['message'] = 'Subscribed!';
} else {
$errors['email'] = 'Error';
}
$conn->close();
// message to me==========================================================================
$to = 'garvernr@mail.uc.edu';
$subject = 'New subscription';
$message = $_POST['email'];
$headers = 'From: newsubscription@samplepackgenerator.com' . "\r\n" .
'Reply-To: newsubscription@samplepackgenerator.com';
mail($to, $subject, $message, $headers);
//message to user=========================================================================
$to = $_POST['email'];
$subject = 'Subscribed!';
$message = 'Hello new member,
Thank you for becoming a part of samplepackgenerator.com. You are now a community member and will recieve light email updates with the lastest information. If you have recieved this email by mistake or wish to no longer be apart of this community please contact nickgarver5@gmail.com
Cheers!,
-Nick Garver ';
$headers = 'From: newsubscription@samplepackgenerator.com' . "\r\n" .
'Reply-To: newsubscription@samplepackgenerator.com';
mail($to, $subject, $message, $headers);
// show a message of success and provide a true success variable==========================
$data['success'] = true;
$data['message'] = 'Subscribed!';
}
?>
HTML
<!-- Subscription -->
<div class="container shorter">
<div class="no-result vertical-align-outer">
<div class="vertical-align">
<form action="phEmail.php" method="POST">
<!-- EMAIL -->
<div id="email-group" class="form-group">
<label for="email"></label>
<input type="text" class="email" name="email" placeholder="Enter your email">
<button type="submit" class="emailbtn">Subscribe</button>
<span></span>
<!-- errors -->
</div>
</div>
</div>
</div>
<br>
<br>
<div id="sub_result">
</div>
答案 0 :(得分：1)
那是因为您忘了编码$data数组。在结束PHP标记（echo json_encode($data);）之前做?>，如下所示：
// your code
mail($to, $subject, $message, $headers);
// show a message of success and provide a true success variable==========================
$data['success'] = true;
$data['message'] = 'Subscribed!';
}
echo json_encode($data);
?>
答案 1 :(得分：1)
你只需要在你的PHP中使用json_encode，因为你的数据类型是json，你期望像json这样的响应
if (!empty($error)){
// your stuff
$data['success'] = false;
$data['errors'] = $errors;
echo json_encode($data);
}
else {
// your stuff
$data['success'] = "SUCCESS MESSAGE";
$data['errors'] = false;
echo json_encode($data);
}
答案 2 :(得分：0)
你的php不会返回任何值，添加一个简单的＆＃34; echo＆＃34;最后一行：
...
$data['success'] = true;
$data['message'] = 'Subscribed!';
}
echo $data['message'];
?>
在js中（如果所有其他代码都正确），您会收到消息。
答案 3 :(得分：0)
您的php文件没有向输出发送任何内容。
添加一行
exit(json_encode($data));
到您想要返回答复的行上的php文件。