Question

我有commentdb.php文件，其中创建了类.like-btn的按钮，并已链接了script.js文件并拥有另一个使用isset($_POST['key'])的db.php文件。

script.js：

$(document).ready(function(){

$('.like-btn').click(function(){

    var post_id = $(this).data('id');
        alert(post_id);
    $clicked_btn = $(this);

    var act = "";

    if ($clicked_btn.hasClass('fa-thumbs-o-up')) {
        act = 'like';
    } else if ($clicked_btn.hasClass('fa-thumbs-up')) {
        act = 'unlike';
    }
    // console.log(act);
    $.ajax({
        type: "POST",
        url: 'db.php',
        data: {
            'key': 1,
            'action': act, 
            'post_id': post_id
        },
        success: function(response) {
            console.log(1);
            // res = JSON.parse(data);
            alert(act);
            if (act == 'like') {
                $clicked_btn.removeClass('fa-thumbs-o-up');
                $clicked_btn.addClass('fa-thumbs-up');
            } else if (act == 'unlike') {
                $clicked_btn.removeClass('fa-thumbs-up');
                $clicked_btn.addClass('fa-thumbs-o-up');
            }
            // alert(res.likes);
    }

    });
});
});

db.php：

<?php  
session_start();
// $email = "$_SESSION['email']";

$conn = mysqli_connect('localhost','root','','commentsection');
echo "working?";
// $user_id = 1;

if (!$conn) {
    die("Connection failed: ".mysqli_connect_error());
}




if (isset($_POST['key'])) {
    echo "isset working?";
    $post_id = $_POST['post_id'];
    $action = $_POST['action'];

    switch ($action) {
        case 'like':
            console.log($email);
            $sql2 = "INSERT INTO rating (uid, rating, user_id) VALUES ($post_id, 'like', $email) ON DUPLICATE KEY UPDATE rating='like'";
            break;
        case 'dislike':
            $sql2 = "INSERT INTO rating (uid, rating, user_id) VALUES ($post_id, 'dislike', $email ON DUPLICATE KEY UPDATE rating='dislike'";
            break;
        case 'unlike':
            $sql2 = "DELETE FROM rating WHERE uid=$post_id AND r_uid=0";
            break;
        case 'undislike':
            $sql2 = "DELETE FROM rating WHERE uid=$post_id AND r_uid=0";
            break;

        default:
            break;
    }


    mysqli_query($conn, $sql2);

    echo getRating($post_id);
    // exit(0);


}

?>

我尝试过使用isset($_POST['key'])，然后使用isset($_POST['action'])和isset($_POST['post_id'])，但是它不起作用，该函数未执行。该条件可能会返回false。

我做了大部分在Google上发现的事情，但没有一个起作用。我真的对此并不了解。因此，请好帮我，坚持一个星期。

使用ajax时，isset（$ _ POST ['']））不适用于数据变量

0 个答案: