试图获取非对象的属性“num_rows”，即使 var_dump 返回一个对象

Question

所以我是 OOP 概念的新手。我决定以面向对象的方式调用我的数据库。我得到的错误是：

试图获取非对象的属性“num_rows”（在第 83 行 *参考第 83 行）

我了解错误消息，但无法找出其中的问题所在。我尝试了以下链接，但不幸的是，它们都没有进一步帮助我，或者我无法理解它们的含义。

Notice: Trying to get property 'num_rows' of non-object in line 35

Error - Trying to get property 'num_rows' of non-object

Get property num_rows of non-object

Trying to get property 'num_rows' of non-object

这就是我故意提出重复问题的原因，希望我的问题是其他帖子中尚未（尚未）解决的问题。

require 'connection.php';


//class DatabaseQueries handles all the queries required regarding CRUD.
class DatabaseQueries
{

    //the SQL string
    private $sql;

    // The connection -> completegallery
    private $conn;



    public function __construct($conn) 
    {

        $this->conn = $conn;

    }


    // function will check whether email already exists or not.
    protected function getEmailExistance(string $email)
    {

        //create the SQL
        $this->sql = $this->conn->prepare("SELECT * FROM userinfo WHERE email = ?");

        //bind the parameter to it (preventing sql injection this way)
        $this->sql->bind_param('s', $email);

        // $result = the execution of the SQL.
        $result = $this->sql->execute();
        $resultCheck = $result->num_rows; //* line 83
        var_dump($result); // returns boolean true.
        var_dump($this->sql); // returns a mysqli stmt object

        //check whether $resultCheck > 0
        //if yes, that would mean the userName already exists.
        if (!empty($result) && $resultCheck > 0) 
        {

            exit('should show something');

        } else 
        {           
            exit('always fails here');

        }
    }

} // ending class DatabaseQueries

我如何称呼 DatabaseQueries 类：

class Base extends DatabaseQueries
        {

            private $email;
            private $userName;
            private $name;
            private $lastName;
            private $pwd;
            private $pwdConfirm;

// here is the code where I check and assign the user input to the variable $email etc. 

            //this method is for test purposes only and will be removed after the website is 'done'.
            public function getEverything()
            {

                //link to check whether email is being used or not
                $this->getEmailExistance($this->email);
            }
}

我如何调用对象等

$userInformation = new base($conn);
$userInformation->setEmail($_POST['registerEmail']);
     //some other info, but not relevant to the problem.

我已经检查过我是否拼错了任何东西，但事实并非如此。 connection.php 中的连接也被声明为正确。

Answer 1

正如 cmets 中提到的，execute() 不会返回结果，您必须 fetch() 它。这是应该允许你做你所要求的代码。

// function will check whether email already exists or not.
protected function getEmailExistance(string $email)
{

    //create the SQL
    $this->sql = $this->conn->prepare("SELECT * FROM userinfo WHERE email = ?");

    //bind the parameter to it (preventing sql injection this way)
    $this->sql->bind_param('s', $email);

    // $result = the execution of the SQL.
    $this->sql->execute(); <---- No need for variable here, its boolean
    $result = $this->sql->store_result(); <---- The result
    $num_rows = $this->sql->num_rows; <---- This will contain your row count

    if (!empty($result)) 
    {
        // fetch your results here
    } else 
    {           
        exit('always fails here');
    }
}