Question

我正在尝试将Rate Yo添加到数据库中。但是我不能总是得到

的错误

未定义索引：C上的评级：\ xampp \ htdocs \ bill \ bill \ add_rate.php 第57行

我尝试了下面所附的代码。

表格

<form action="add_rate.php" method="post">

    <div class="rateyo" id= "rating"
         data-rateyo-rating="4"
         data-rateyo-num-stars="5"
         data-rateyo-score="3"></div>
    <span class='score'>0</span>
    <span class='result'>0</span>
    </div>

    <div><input type="submit" name="add"> </div>
</form>

jQuery

$(function () {
        $(".rateyo").rateYo().on("rateyo.change", function (e, data) {
            var rating = data.rating;
            $(this).parent().find('.score').text('score :'+ 
        $(this).attr('data-rateyo-score'));
            $(this).parent().find('.result').text('rating :'+ rating);
        });
    });

php代码

require 'db_connection.php';
if ($_SERVER["REQUEST_METHOD"] == "POST")
{
    $rating = $_POST["rating"];
    $conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
    if (!$conn)
    {
        die("Connection failed: " . mysqli_connect_error());
    }
    $sql = "INSERT INTO ratee (rate) VALUES ('$rating')";
    if (mysqli_query($conn, $sql))
    {
        echo "New Rate addedddd successfully";
    }
    else
    {
        echo "Error: " . $sql . "<br>" . mysqli_error($conn);
    }
    mysqli_close($conn);

}
?>

Answer 1

如下所示更改表单代码和jquery代码：

<form action="add_rate.php" method="post">

    <div class="rateyo" id= "rating"
         data-rateyo-rating="4"
         data-rateyo-num-stars="5"
         data-rateyo-score="3"></div>
    <span class='score'>0</span>
    <span class='result'>0</span>
    <input type="text" name="rating">
    </div>

    <div><input type="submit" name="add"> </div> <!-- added hidden field with name rating
</form>

jquery代码：-

$(function () {
    $(".rateyo").rateYo().on("rateyo.change", function (e, data) {
        var rating = data.rating;
        $(this).parent().find('.score').text('score :'+ $(this).attr('data-rateyo-score'));
        $(this).parent().find('.result').text('rating :'+ rating);
        $(this).parent().find('input[name=rating]').val(rating); //add rating value to input field
    });
});

稍微更改php代码：-

if (!empty($_POST["rating"]))
{
    //rest of your code
}else{
   die('ratings required');
}

注意：-您的php代码已经可以进行SQL注入，因此请使用prepared statements来阻止它。

如何将Rate Yo添加到数据库php mysqli中

1 个答案: