Question

也许这是一个可能的重复，我在这里搜索一些这样的问题，但我已经尝试了所有的答案，但我仍然有这个错误

Notice: Undefined property: User::$_pdo in D:\xampp\htdocs\pengun\classes\DB.php on line 32

和此错误

Fatal error: Call to a member function prepare() on a non-object in D:\xampp\htdocs\pengun\classes\DB.php on line 32

这是我的DB类

<?php
class DB {

    private $_pdo, $_query, $_result, $_count, $_row;

    public function __construct($host, $dbname, $user, $password) {
        $this->host = $host;
        $this->user = $user;
        $this->password = $password;
        $this->dbname = $dbname;

        try {
            $_pdo = new PDO('mysql:host=' . $this->host . ';dbname=' . $this->dbname, $this->user, $this->password);
        } catch (PDOException $e) {
            die($e->getMessage());
        }

        $_pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
    }

    public function select($fields, $table, $where = array()) {
        if(count($where) === 3) {
            $operators = array('=','>','<','>=','<=');

            $column     = $where[0];
            $operator   = $where[1];
            $value      = $where[2];

            if(in_array($operator, $operators)) {
                $sql = "SELECT {$fields} FROM {$table} WHERE {$column} {$operator} {$value}";
                if($this->_query = $this->_pdo->prepare($sql)) {
                    $this->_query->execute();
                    $this->_row = $this->_query->fetch();

                    print_r($this->_row);
                }
            }
        }
    }
}

有人可以告诉我我的代码有什么问题吗？提前谢谢。

Answer 1

因为在构造函数中，您只将new PDO分配给只能在构造函数内部访问的局部变量。你必须使用：

$this->_pdo = new PDO('mysql:host=' . $this->host . ';dbname=' . $this->dbname, $this->user, $this->password);
...
$this->_pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

Answer 2

您在构造函数中为PDO对象分配了局部变量。你必须设置类变量！

更改

$_pdo

到

$this->_pdo

第13和18行。

php中的类undefined属性

2 个答案: