要使用jQuery发出Ajax请求，可以通过以下代码实现。

HTML:

<form id="foo">
    <label for="bar">A bar</label>
    <input id="bar" name="bar" type="text" value="" />
    <input type="submit" value="Send" />
</form>

<!-- The result of the search will be rendered inside this div -->
<div id="result"></div>

JavaScript:

方法1

 /* Get from elements values */
 var values = $(this).serialize();

 $.ajax({
        url: "test.php",
        type: "post",
        data: values ,
        success: function (response) {

           // You will get response from your PHP page (what you echo or print)
        },
        error: function(jqXHR, textStatus, errorThrown) {
           console.log(textStatus, errorThrown);
        }
    });

方法2

/* Attach a submit handler to the form */
$("#foo").submit(function(event) {
    var ajaxRequest;

    /* Stop form from submitting normally */
    event.preventDefault();

    /* Clear result div*/
    $("#result").html('');

    /* Get from elements values */
    var values = $(this).serialize();

    /* Send the data using post and put the results in a div. */
    /* I am not aborting the previous request, because it's an
       asynchronous request, meaning once it's sent it's out
       there. But in case you want to abort it you can do it
       by abort(). jQuery Ajax methods return an XMLHttpRequest
       object, so you can just use abort(). */
       ajaxRequest= $.ajax({
            url: "test.php",
            type: "post",
            data: values
        });

    /*  Request can be aborted by ajaxRequest.abort() */

    ajaxRequest.done(function (response, textStatus, jqXHR){

         // Show successfully for submit message
         $("#result").html('Submitted successfully');
    });

    /* On failure of request this function will be called  */
    ajaxRequest.fail(function (){

        // Show error
        $("#result").html('There is error while submit');
    });

从jQuery 1.8开始，.success()、.error()和.complete()回调函数已弃用。要为最终删除它们做好准备，请使用.done()、.fail()和.always()来代替。

MDN: abort()。如果请求已经发送，此方法将中止请求。

我们已经成功地发送了一个Ajax请求，现在是时候向服务器抓取数据了。

PHP

当我们在Ajax调用中发出POST请求(类型:" POST ")时，我们现在可以使用$_REQUEST或$_POST获取数据:

  $bar = $_POST['bar']

您还可以通过以下两种方式查看您在POST请求中获得了什么。顺便说一句，确保设置了$_POST。否则您将得到一个错误。

var_dump($_POST);
// Or
print_r($_POST);

您正在向数据库中插入一个值。确保在执行查询之前正确地敏感或转义了所有请求(无论是GET还是POST)。最好的方法是使用准备好的语句。

如果你想将任何数据返回到页面，你可以通过如下所示的回显数据来实现。

// 1. Without JSON
   echo "Hello, this is one"

// 2. By JSON. Then here is where I want to send a value back to the success of the Ajax below
echo json_encode(array('returned_val' => 'yoho'));

然后你可以得到:

 ajaxRequest.done(function (response){
    alert(response);
 });

有几种简便的方法。您可以使用下面的代码。它做同样的功。

var ajaxRequest= $.post("test.php", values, function(data) {
  alert(data);
})
  .fail(function() {
    alert("error");
  })
  .always(function() {
    alert("finished");
});

你可以使用serialize。下面是一个例子。

$("#submit_btn").click(function(){
    $('.error_status').html();
        if($("form#frm_message_board").valid())
        {
            $.ajax({
                type: "POST",
                url: "<?php echo site_url('message_board/add');?>",
                data: $('#frm_message_board').serialize(),
                success: function(msg) {
                    var msg = $.parseJSON(msg);
                    if(msg.success=='yes')
                    {
                        return true;
                    }
                    else
                    {
                        alert('Server error');
                        return false;
                    }
                }
            });
        }
        return false;
    });

我使用如下所示的方法。它像文件一样提交所有东西。

$(document).on("submit", "form", function(event)
{
    event.preventDefault();

    var url  = $(this).attr("action");
    $.ajax({
        url: url,
        type: 'POST',
        dataType: "JSON",
        data: new FormData(this),
        processData: false,
        contentType: false,
        success: function (data, status)
        {

        },
        error: function (xhr, desc, err)
        {
            console.log("error");
        }
    });
});

自从引入了Fetch API，就没有理由再用jQuery Ajax或xmlhttprequest来实现了。要POST表单数据到php脚本在香草JavaScript你可以做以下:

异步函数postData() { 尝试{ Const res = await fetch('../php/contact.php'， { 方法:“文章”, new FormData(document.getElementById('form')) }) if (!res.ok)抛出新的错误('网络响应不正常'); } catch (err) { console.log (err) ｝ ｝ <form id="form" action="javascript:postData()" > <input id="name" name="name" placeholder=" name" type="text" required> <input type="submit" value=" submit" > > < /形式

下面是一个非常基本的php脚本示例，它获取数据并发送电子邮件:

<?php
    header('Content-type: text/html; charset=utf-8');

    if (isset($_POST['name'])) {
        $name = $_POST['name'];
    }

    $to = "test@example.com";
    $subject = "New name submitted";
    $body = "You received the following name: $name";
    
    mail($to, $subject, $body);

在你的php文件中输入:

$content_raw = file_get_contents("php://input"); // THIS IS WHAT YOU NEED
$decoded_data = json_decode($content_raw, true); // THIS IS WHAT YOU NEED
$bar = $decoded_data['bar']; // THIS IS WHAT YOU NEED
$time = $decoded_data['time'];
$hash = $decoded_data['hash'];
echo "You have sent a POST request containing the bar variable with the value $bar";

在你的js文件中发送一个带有数据对象的ajax

var data = { 
    bar : 'bar value',
    time: calculatedTimeStamp,
    hash: calculatedHash,
    uid: userID,
    sid: sessionID,
    iid: itemID
};

$.ajax({
    method: 'POST',
    crossDomain: true,
    dataType: 'json',
    crossOrigin: true,
    async: true,
    contentType: 'application/json',
    data: data,
    headers: {
        'Access-Control-Allow-Methods': '*',
        "Access-Control-Allow-Credentials": true,
        "Access-Control-Allow-Headers" : "Access-Control-Allow-Headers, Origin, X-Requested-With, Content-Type, Accept, Authorization",
        "Access-Control-Allow-Origin": "*",
        "Control-Allow-Origin": "*",
        "cache-control": "no-cache",
        'Content-Type': 'application/json'
    },
    url: 'https://yoururl.com/somephpfile.php',
    success: function(response){
        console.log("Respond was: ", response);
    },
    error: function (request, status, error) {
        console.log("There was an error: ", request.responseText);
    }
  })

或者保持表单提交的原样。只有当您希望发送带有计算附加内容的修改过的请求，而不仅仅是一些由客户机输入的表单数据时，才需要这种方法。例如哈希、时间戳、用户id、会话id等等。

我想分享一个详细的方法，如何张贴与PHP + Ajax以及错误抛出返回失败。

首先，创建两个文件，例如form.php和process.php。

我们将首先创建一个表单，然后使用jQuery .ajax()方法提交。其余的将在评论中解释。

form.php

<form method="post" name="postForm">
    <ul>
        <li>
            <label>Name</label>
            <input type="text" name="name" id="name" placeholder="Bruce Wayne">
            <span class="throw_error"></span>
            <span id="success"></span>
       </li>
   </ul>
   <input type="submit" value="Send" />
</form>

使用jQuery客户端验证表单并将数据传递给process.php。

$(document).ready(function() {
    $('form').submit(function(event) { //Trigger on form submit
        $('#name + .throw_error').empty(); //Clear the messages first
        $('#success').empty();

        //Validate fields if required using jQuery

        var postForm = { //Fetch form data
            'name'     : $('input[name=name]').val() //Store name fields value
        };

        $.ajax({ //Process the form using $.ajax()
            type      : 'POST', //Method type
            url       : 'process.php', //Your form processing file URL
            data      : postForm, //Forms name
            dataType  : 'json',
            success   : function(data) {
                            if (!data.success) { //If fails
                                if (data.errors.name) { //Returned if any error from process.php
                                    $('.throw_error').fadeIn(1000).html(data.errors.name); //Throw relevant error
                                }
                            }
                            else {
                                    $('#success').fadeIn(1000).append('<p>' + data.posted + '</p>'); //If successful, than throw a success message
                                }
                            }
        });
        event.preventDefault(); //Prevent the default submit
    });
});

现在我们来看看process.php

$errors = array(); //To store errors
$form_data = array(); //Pass back the data to `form.php`

/* Validate the form on the server side */
if (empty($_POST['name'])) { //Name cannot be empty
    $errors['name'] = 'Name cannot be blank';
}

if (!empty($errors)) { //If errors in validation
    $form_data['success'] = false;
    $form_data['errors']  = $errors;
}
else { //If not, process the form, and return true on success
    $form_data['success'] = true;
    $form_data['posted'] = 'Data Was Posted Successfully';
}

//Return the data back to form.php
echo json_encode($form_data);

项目文件可以从http://projects.decodingweb.com/simple_ajax_form.zip下载。