PHP MySQL在预准备语句中围绕匿名变量抛出错误

Question

当我运行我准备好的语句时,我收到以下错误:

错误:INSERT INTO文章(url,headline,pubDate,source,image_loc(？？？？？,,,,))VALUES您的SQL语法错误; 查看与您的MySQL服务器版本对应的手册,以便在第6行使用"？,？,？,？,？)"附近的正确语法

这是似乎抛出错误的代码:

            $sql = "INSERT INTO $tableName (`url`, 
                            `headline`, 
                            `pubDate`, 
                            `source`, 
                            `image_loc`) 
                VALUES(?, ?, ?, ?, ?)";

        // MySQLi connection, binds variables to prevent injection, executes
        $stmt = $connection->prepare($sql);
        $stmt->bind_param('sssss', $url, $headline, $pubDate, $source, $image_loc);
        $stmt->execute();

编辑:这是我在单独的文件中设置为我的连接.它起作用的意义是一切都被保存了......我只是在抛出一个错误.

$servername = "localhost";
$username = "xxxxxx";
$password = "xxxxxx";
$dbname = "news";
$tableName = "articles";

$connection = mysqli_connect($servername, $username, $password, $dbname);

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

再次编辑:这是用来检查插入成功的代码,虽然我猜这可能是问题的根源(但我没有足够的经验来理解为什么):

if (mysqli_query($connection, $sql)) {
            echo "New record created successfully";
        } else {
            echo "Error: " . $sql . "<br>" . mysqli_error($connection) . "<br>";
        }

Answer 1

你的代码:

if (mysqli_query($connection, $sql)) {
    echo "New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . mysqli_error($connection) . "<br>";
}

......用途mysqli_query().这实际上是执行查询的另一种方式,只是不同于execute()它不使用绑定参数.它只是将查询发送到数据库,问号和所有,从而产生您的错误.

如果要检查错误,应execute()在运行时检查语句的结果,而不是执行进一步的查询.$stmt->execute()用这样的东西替换你的行,执行查询并测试成功:

if ($stmt->execute()) {
    // Success
} else {
    // Failure
}