未定义变量:pdo，在 null 上调用成员函数 prepare()

Undefined variable: pdo, call to a member function prepare() on null(未定义变量:pdo，在 null 上调用成员函数 prepare())

问题描述

我正在关注一个视频并仔细检查所有代码，一切似乎都一样，但我遇到了这些错误.

I was following a video and double checking all code and everything seems to be the same yet I get these errors.

错误:

注意:未定义变量:第 14 行 QueryBuilder.php 中的 pdo

致命错误:在 QueryBuilder.php 第 14 行调用成员函数 prepare() 为 null

QueryBuilder.php:

class QueryBuilder
{
    protected $pdo;

    public function __construct($pdo)
    {
        $this->pdo = $pdo;
    }

    public function selectAll($table)
    {
        $query = $pdo->prepare("SELECT * FROM `$table`"); // --> LINE 14 <--
        $query->execute();
        return $query->fetchAll();
    }
}

Connection.php:

class Connection
{
    public static function make()
    {
        $servername = "localhost";
        $dbUsername = "root";
        $dbPassword = "";
        $dbName = "test";

        try {
            $pdo = new PDO("mysql:host=$servername;dbname=$dbName", $dbUsername, $dbPassword);
            $pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
            $pdo->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
            return $pdo;
        }
        catch(PDOException $e){
            die($e->getMessage());
        }
    }
}

init.php:

require "database/Connection.php";
require "database/QueryBuilder.php";
require "app/Product.php";

$query = new QueryBuilder(Connection::make());

推荐答案

如评论中所述，在 OOP 中，您需要使用 $this->pdo 为其传递对象的属性，而不是变量 $query = $pdo-> ，因为你已经在:

As stated in comments, in OOP, you need to use $this->pdo passing the object's property for it, instead of the variable $query = $pdo-> since you've construct'ed it in:

public function __construct($pdo)
{
    $this->pdo = $pdo;
    ^^^^^^^^^^
}

即:

$query = $this->pdo->prepare

这篇关于未定义变量:pdo，在 null 上调用成员函数 prepare()的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持编程学习网！