Question

我明白了：可捕获的致命错误：当我想将数据插入到我的游戏桌中时，类mysqli的对象无法转换为字符串错误。在我的games.php上，我有一个将数据发送到games_add.php的表单，并且在games_add.php的第18行发生错误。

（我知道很多代码）game.php代码：

<body>
  <form action="games_add.php" method="post">
Game name: <input type="text" name="game_name" placeholder="Enter first name ..." required="required" /><br />
Relese date: <input type="date" name="relese_date" placeholder="Enter  relese date ..." required="required" /><br />
Introduction: <input type="text" name="introduction" placeholder="Enter introduction ..."  required="required" /><br />
Description: <input type="text" name="description" placeholder="Enter description ..."  required="required" /><br />
Original price: <input type="number" name="rating" placeholder="Enter original price ..." required="required" /><br />

Developer: <select name="developer_id">
    <?php 
    include_once 'connection.php';
        $query = "SELECT * FROM developers";
        $result = mysqli_query($link, $query);
        while ($row = mysqli_fetch_array($result)) {
             echo '<option value="'.$row['id'].'">'.$row['name'].'</option>';
        }     
    ?>
</select>    

    Publisher: <select name="publisher_id">
    <?php 
    include_once 'connection.php';
        $query = "SELECT * FROM publishers";
        $result = mysqli_query($link, $query);
        while ($row = mysqli_fetch_array($result)) {
             echo '<option value="'.$row['id'].'">'.$row['name'].'</option>';
        }     
    ?>
</select> 



    Categories: <select name="categories_id">
    <?php 
    include_once 'connection.php';
        $query = "SELECT * FROM categories";
        $result = mysqli_query($link, $query);
        while ($row = mysqli_fetch_array($result)) {
             echo '<option value="'.$row['id'].'">'.$row['name'].'</option>';
        }     
    ?>
</select>



    Platform: <select name="platform1_id">
    <?php 
    include_once 'connection.php';
        $query = "SELECT * FROM platforms";
        $result = mysqli_query($link, $query);
        while ($row = mysqli_fetch_array($result)) {
             echo '<option value="'.$row['id'].'">'.$row['name'].'</option>';
        }     
    ?>
</select>


<input type="submit" value="Insert" />
<input type="reset" value="Cancle" />

games_add.php代码：

      <?php
    include_once 'connection.php';

        $game_name = $_POST['game_name'];
        $relese_date = $_POST['relese_date'];
        $introduction = $_POST['introduction'];
        $description = $_POST['description'];
        $rating = $_POST['rating'];
        $developer_id = $_POST['developer_id'];
        $publisher_id = $_POST['publisher_id'];
        $categories_id = $_POST['categories_id'];
        $platform1_id = $_POST['platform1_id'];
        $avrage_score = 33;



            $query = sprintf("INSERT INTO games (game_name, relese_date, introduction, description, rating, developer_id, publisher_id, categories_id, platform1_id, avrage_score ) 
                              VALUES ('%s','$relese_date','%s','%s','$rating','%s','%s','%s','%s','$avrage_score') or die(mysqli_error($link)); ",
                             mysqli_real_escape_string($link, $game_name),
                             mysqli_real_escape_string($link, $introduction),
                             mysqli_real_escape_string($link, $description),
                             mysqli_real_escape_string($link, $developer_id),
                             mysqli_real_escape_string($link, $publisher_id),
                             mysqli_real_escape_string($link, $categories_id),
                             mysqli_real_escape_string($link, $platform1_id));                
            mysqli_query($link, $query);
 header('Location: games.php');
?>

我的数据库的图片是否有帮助。

Answer 1

你确定你想要＆＃34;或死（mysqli_error（$ link））＆＃34;成为SQL的一部分，而不是PHP的一部分？

插入表格时出错

1 个答案: