PHP类对象内部方法

Question

我在使用在类的第二个方法中的类的一个方法中实例化的mysqli对象时遇到麻烦。 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();
}

我认为我的麻烦是将实例化的对象公开。

****更新：这是我目前拥有的整个类，并且在尝试执行查询方法中的一行时脚本快死了：

$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.

Answer 1

您的问题是变量作用域。 如果不global指定，则无法以所需方式访问它。

您最好坚持最佳实践并执行以下操作：

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';
    }

}

您会注意到我们如何将其定义为$this （在对象内）。 现在您可以访问它：

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();
}

Answer 2

方法是一种功能。 与其他任何函数一样，在函数内部声明的变量无法在该函数外部访问。