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如何让php页面从html页面接收ajax帖子

[英]How to get php page to receive ajax post from html page

I have a very simple form that has an input field for first name.我有一个非常简单的表单,它有一个名字的输入字段。 I captured the form data and transmitted it via ajax to a PHP page using the standard jQuery posting method.我捕获了表单数据并使用标准的 jQuery 发布方法通过 ajax 将其传输到 PHP 页面。 However, I am not able at all get any responses from the PHP page that any data was captured on the server-side.但是,我根本无法从 PHP 页面获得任何响应,即在服务器端捕获了任何数据。 I am not sure what I have done wrong or what is missing.我不确定我做错了什么或遗漏了什么。

Here is my code.这是我的代码。

Form:形式:

<form action="process.php" method="POST">
<div class="form-group">
<div class="form-row">
<div class="col-md-6 mb-3">
<label for="firstName">First name</label>
<input type="text" class="form-control" name="firstName" id="firstName" placeholder="First name">
<div class="d-none" id="firstName_feedback">
<p>Please enter a first name.</p>
</div>
</div>
</div>
</div>
<button class="btn btn-primary" type="submit">Submit form</button>
</form>

Here is my Jquery Ajax call:这是我的 Jquery Ajax 调用:

<script>
$(document).ready(function() {
$('form').submit(function(event) {
var formData = $("form").serialize();
console.log(formData); 
$.ajax({
type: 'POST', 
url: 'form.php', 
data: formData, 
dataType: 'json', 
encode: true    
   })
.done(function(data) {
 console.log(data); 
 });
 event.preventDefault();
 });

 });


</script>

And here is my PHP page:这是我的 PHP 页面:

if(isset($_POST['formData']))    
$ajaxData = ($_POST['formData']); 
echo $ajaxData;
{
}

In your Ajax function, you're passing the contents of formData to the server, though not as formData but as their original input name.在您的 Ajax 函数中,您将formData的内容formData给服务器,尽管不是作为formData而是作为它们的原始输入名称。

In this case, you have:在这种情况下,您有:

<input type="text" class="form-control" name="firstName" id="firstName" placeholder="First name">

The input's name is firstName, so you need to call $_POST['firstName'] instead of $_POST['formData'] .输入的名称是 firstName,因此您需要调用$_POST['firstName']而不是$_POST['formData']

if (isset($_POST['firstName'])) {
    $ajaxData = $_POST['firstName'];
    echo $ajaxData;
}

The same applies for any other field you would have in your form, so for example, having another input with the name lastName means you'd have to call $_POST['lastName'] to access it.这同样适用于表单中的任何其他字段,例如,具有名称为lastName另一个输入意味着您必须调用$_POST['lastName']来访问它。

There were also some misplaced brackets and parentheses in the PHP code which I accommodated above.我上面提到的 PHP 代码中也有一些错位的方括号和圆括号。

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