如何在 PHP 中处理从 jQuery 序列化并发送的数据？

发布于 2024-11-07 13:04:22 字数 1625 浏览 0 评论 0原文

我正在努力获取 $.ajax 调用以正确地将一些表单数据发送到 PHP（将其记录到数据库然后显示）。我很困惑，因为我之前用 $.ajax 来完成类似的任务，而且效果很好，但我一定在这里遗漏了一些关键的东西。我研究了其他答案（例如这个），但找不到任何表明我的答案的内容当前的代码将无法工作。任何见解将不胜感激！

表单如下所示：

            <div id="note_add_container">
                <form id="note_add" method="POST">
                    <input type="text" name="title" placeholder="title" />
                    <input type="text" name="summary" placeholder="summary" />
                    <input type="text" name="details" placeholder="details" />
                    <button id="submit_note">Add note!</button>
                </form>
            </div>

            <div id="entries">
                <!-- AJAX call will populate entries here -->
            </div>

这是 jQuery：

    $('#submit_note').click(function () {
        var text = $.ajax ({
            type: "POST",
            url: "note_process.php",
            data: $('#note_add').serialize(), 
            dataType: "json",
            async: false,
        }).responseText;
        $('#entries').html(text); 
    })

这是 PHP note_process.php：

include_once "connect.php";
session_start();

$id = $_SESSION['userid'];
$title = $_POST['title'];
$summary = $_POST['summary'];
$details = $_POST['details'];

$query = mysql_query("INSERT INTO notes (id, title, summary, details) VALUES ('$id', '$title','$summary','$details')");

echo $title . $summary . $details;

原文

I'm struggling to get a $.ajax call to correctly send some form data to PHP (where it gets recorded to a database and then displayed). I'm stumped because I've $.ajax to do similar tasks before and it's worked great but I must be missing something critical here. I've researched other answers (such as this one) but can't find anything there that suggests my current code would not work. Any insight would be greatly appreciated!

The form looks like this:

            <div id="note_add_container">
                <form id="note_add" method="POST">
                    <input type="text" name="title" placeholder="title" />
                    <input type="text" name="summary" placeholder="summary" />
                    <input type="text" name="details" placeholder="details" />
                    <button id="submit_note">Add note!</button>
                </form>
            </div>

            <div id="entries">
                <!-- AJAX call will populate entries here -->
            </div>

Here is the jQuery:

    $('#submit_note').click(function () {
        var text = $.ajax ({
            type: "POST",
            url: "note_process.php",
            data: $('#note_add').serialize(), 
            dataType: "json",
            async: false,
        }).responseText;
        $('#entries').html(text); 
    })

Here is the PHP note_process.php:

include_once "connect.php";
session_start();

$id = $_SESSION['userid'];
$title = $_POST['title'];
$summary = $_POST['summary'];
$details = $_POST['details'];

$query = mysql_query("INSERT INTO notes (id, title, summary, details) VALUES ('$id', '$title','$summary','$details')");

echo $title . $summary . $details;

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宫墨修音 2024-11-14 13:04:22

尝试使用“成功”，而不是使 Ajax 请求非异步...

   $('#submit_note').click(function () {
        $.ajax ({
            type: "POST",
            url: "note_process.php",
            data: $('#note_add').serialize(), 
            success: function(text){
                  $('#entries').html(text); 
            }
        });    
    });

Try using 'success', instead of making the Ajax request not-async...

   $('#submit_note').click(function () {
        $.ajax ({
            type: "POST",
            url: "note_process.php",
            data: $('#note_add').serialize(), 
            success: function(text){
                  $('#entries').html(text); 
            }
        });    
    });

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