php - 无法显示 SELECT 结果 sql

Question

我尝试获取 gmail = $gmail 的 id 并将其用于 UPDATE 表

但我不能使用 from id 和 showin 错误未定义属性:mysqli_result::$fetch_assoc

函数添加验证码($gmail , $random) {

    $connection = mysqli_connect(DataBaseManager::HOST, DataBaseManager::USER, DataBaseManager::PASSWORD, DataBaseManager::DATABASENAME);
    mysqli_set_charset($connection, "utf8");
    $sqlQuery = "SELECT id FROM users WHERE gmail='$gmail'";
    $result = mysqli_query($connection, $sqlQuery);
    if ($result->num_rows > 0) {
            $sqlCommand = "UPDATE users
                      SET verificationcode  = '$random' 
                        WHERE id = $result->fetch_assoc()";
    }

    if(mysqli_query($connection, $sqlCommand)){
        return true;
    }else{
        echo("Error description: " . mysqli_error($connection));
        return false;
    }


    }

Answer 1

你必须使用 curly braces在对象中调用方法时:

        $sqlCommand = "UPDATE users
                  SET verificationcode  = '$random' 
                    WHERE id = {$result->fetch_assoc()}";

另一种方法是使用连接:

        $sqlCommand = "UPDATE users
                  SET verificationcode  = '$random' 
                    WHERE id = " . $result->fetch_assoc();

请注意，在这种情况下，您可以将两个 SQL 语句合并为一个，例如 update .. where id = (select id from ...)。

另请注意，您发布的代码容易受到 SQL 注入(inject)攻击。参见 How can I prevent SQL injection in PHP?