PHP 类对象在方法中 - PHP Class Object inside method

PHP Class Object inside method

本文关键字：方法对象 PHP | 更新日期: 2023-09-27

我在使用该类的第二个方法中在我的类的一个方法中实例化的 mysqli 对象时遇到问题。谁能帮助我了解我可能做错了什么？这是我的两个类方法的样子：

function connect() {
    $mysqli = new mysqli($this->host, $this->user, $this->pass, $this->db);
    if($mysqli->connect_errno) {
        printf("Connect failed: %s'n", $mysqli->connect_error);
        exit();
    }
    else {
        echo 'Successful';
    }
}
public function query($sql) {
    $this->sql = $sql;
    $mysqli->connect();
    $result = $mysqli->query($this->sql);
    while ($row = $result->fetch_assoc()) {
        printf ("$s (%s)'n", $row['name']);
        echo "<br>";
    }
    $result->free();
    $mysqli->close();
}

我认为我的麻烦是公开实例化的对象。

更新：这是当前我拥有的整个类，当尝试在查询方法中执行一行时，脚本正在死亡：

$this->mysqli->connect();

class database {
    public $host = "host";
    public $user = "user";
    public $pass = "pw";
    public $db = "dbname";
    public $sql;
    private $mysqli;
    function __construct() {
        echo "new class object created";
        echo "<br><br>";
    }
    function connect() {
        $this->mysqli = new mysqli($this->host, $this->user, $this->pass, $this->db);
        if($this->mysqli->connect_errno) {
            printf("Connect failed: %s'n", $mysqli->connect_error);
            exit();
        }
        else {
            echo 'Successful';
        }
    }
    public function query($sql) {
        $this->sql = $sql;
        echo "test1";
        $this->mysqli->connect();
        echo "test2";
        $result = $this->mysqli->query($this->sql);
        echo "test3";
        while ($row = $result->fetch_assoc()) {
            printf ("$s (%s)'n", $row['name']);
            echo "<br>";
        }
        $result->free();
        $this->mysqli->close();
    }
}

echo "test2"; does not execute.

你的问题是变量范围。如果不global 'ly 指定它，则无法以您想要的方式访问它。

你最好坚持最佳实践并做这样的事情：

private $mysqli;
function connect() {
    $this->mysqli = new mysqli($this->host, $this->user, $this->pass, $this->db);
    if($this->mysqli->connect_errno) {
        printf("Connect failed: %s'n", $mysqli->connect_error);
        exit();
    }
    else {
        echo 'Successful';
    }
}

您注意到我们如何将其定义为$this（在对象内）。现在您将能够访问它：

public function query($sql) {
    $this->sql = $sql;
    $this->connect();
    $result = $this->mysqli->query($this->sql);
    while ($row = $result->fetch_assoc()) {
        printf ("$s (%s)'n", $row['name']);
        echo "<br>";
    }
    $result->free();
    $this->mysqli->close();
}

方法是一个函数。就像任何其他函数一样，您在函数内声明的变量不能在该函数外部访问。