使用AJAX调用PHP文件

我一直在研究其他问题，并根据相应的答案编写代码，但我似乎无法做到这一点。如果有比我更有经验的人能帮我看得更清楚，我将不胜感激

jQuery:

$(function () {
$(".going-decision").click(function () {
    var clickBtnValue = $(this).val();
    alert('clicked');
    $.ajax({
        type: "POST",
        url: "http://localhost/townbuddies/public_html/event_going_ajax.php",
        data: {'btn-clicked': clickBtnValue},
        success: function () {
            alert('success');
        }
    });
});
});

if (isset($_POST['btn-clicked'])) {
    $decision = $_POST['btn-clicked'];
    //Logic that writes to database
}

event\u going\u ajax.php:

$(function () {
$(".going-decision").click(function () {
    var clickBtnValue = $(this).val();
    alert('clicked');
    $.ajax({
        type: "POST",
        url: "http://localhost/townbuddies/public_html/event_going_ajax.php",
        data: {'btn-clicked': clickBtnValue},
        success: function () {
            alert('success');
        }
    });
});
});

if (isset($_POST['btn-clicked'])) {
    $decision = $_POST['btn-clicked'];
    //Logic that writes to database
}

我的HTML按钮确实有“going decision”类。包括jQuery库

结果：将显示“单击”警报以及“成功”警报。但是，数据库没有更新

有人知道这里发生了什么吗？谢谢

编辑，这是我的PHP代码。我知道这是不安全的代码，我目前正在学习PHP。

<?php

session_start();
$user_id = $_SESSION['user'];

define('DB_HOST', 'localhost');
define('DB_USER', 'root');
define('DB_PASSWORD', '');
define('DB_NAME', 'townbuddies');

$con = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);

if (isset($_POST['btn-clicked'])) {
//Value will be in this format: 'decision.event_id'
$value = $_POST['btn-clicked'];
$length = strlen($value);
$findme = '.';
$pos = strpos($value, $findme);

//Separating the decision (going/not going) and the event_id
$decision = substr($value, 0, $pos);
$event_id = substr($value, $pos + 1);

//Verify if user is already 'going' to the event
$result = verifyDatabase($con, $user_id, $event_id);

//Depending on user decision
switch ($decision) {
    case 'going':
        going($con, $result, $user_id, $event_id);
        break;
    case 'not_going':
        notGoing($con, $result, $user_id, $event_id);
        break;
}
}

function verifyDatabase($con, $user_id, $event_id) {
//Verify if already in database
$sql = "SELECT * "
        . "FROM user_events "
        . "WHERE event_id = "
        . "'$event_id'"
        . "AND user_id = "
        . "'$user_id'";

$result = mysqli_query($con, $sql);

return $result;
}

function going($con, $result, $user_id, $event_id) {
//If already in database as 'going', then do not duplicate. Do nothing.
if (!empty($result)) {
    header("Location: http://localhost/townbuddies/public_html/event.php?event=" . $event_id);
    exit;
}

//Not in database as 'going', so add it.
else {
    $sql = "INSERT INTO user_events (user_id,event_id) "
            . "VALUES ("
            . "'$user_id'"
            . ","
            . "'$event_id'"
            . ")";

    if (!mysqli_query($con, $sql)) {
        echo "Error: " . $sql . "<br>" . mysqli_error($con);
    }

    //Update amount of people going.
    $sql = 

    mysqli_close($con);
}
exit;
}

function notGoing($con, $result, $user_id, $event_id) {
//If already in database as 'going', then delete
if (!empty($result)) {
    $sql = "DELETE FROM user_events "
            . "WHERE user_id = "
            . "'$user_id'"
            . "AND event_id = "
            . "'$event_id'";

    if (!mysqli_query($con, $sql)) {
        echo "Error: " . $sql . "<br>" . mysqli_error($con);
    }

    mysqli_close($con);
}
exit;
}

?>

将无法按预期工作。。。如果我交换If和else语句，一切正常

---

为什么mySQL的空结果对于该语句来说不是空的？

我希望我是错的，但是您的SQL查询有一个错误

$sql = "INSERT INTO user_events (user_id,event_id) "
            . "VALUES ("
            . "'$user_id'"
            . ","
            . "'$event_id'"; // should be ."'$event_id')"

---

条件没有结束括号，这将中断查询。

If

alert（'success'），那么您的问题是php文件
事件_going_ajax.php
，而不是您的js代码。您能在控制台上看到请求吗？（F12）？？我可以看看你的
事件吗\u go\u ajax.php
代码？你能检查一下你的错误日志并告诉我们你得到了什么吗。如果你得到一个错误？如果你不明白我在说什么，你有一个更大的问题@LeaTano我可以在“网络”选项卡下看到请求。我将在几秒钟内编辑我的原始帖子，以显示所有php代码。非常感谢。编辑我的代码以包含PHP。正如
mysqli_query
文档所说，“失败时返回FALSE。对于成功的选择、显示、描述或解释查询，mysqli_query（）将返回mysqli_结果对象。对于其他成功的查询，mysqli_query（）将返回TRUE。”因此，这就是它总是返回TRUE或FALSE的原因。谢谢。这并没有解决我遇到的问题，但肯定是我代码中的一个问题。