Question

我有连接的配置文件但是当我使用$ this-＆gt; con＆gt; prepare（）或$ this-＆gt; con＆gt; query（）不起作用。但是，当我为连接创建单独的变量时，下面的工作很好，例如。

此代码无效

private function isEmailexist($email)
{
    $stmt = $this->con->prepare("SELECT * FROM devices WHERE email = ?");
    $stmt->bind_param("s",$email);
    $stmt->execute();
    $stmt->store_result();
    $rows = $stmt->num_rows;
    echo "Rows : ".$rows;
}

此代码工作

private function isEmailexist($email)
{
    $mysqli = new mysqli("localhost", "root", "", "demo1");
    $stmt = $mysqli->prepare("SELECT * FROM devices WHERE email = ?");
    $stmt->bind_param("s",$email);
    $stmt->execute();
    $stmt->store_result();
    $rows = $stmt->num_rows;
    echo "Rows : ".$rows;
}

这是配置文件

class Config
{
    private $con;
    function __construct(){}
    function connect()
    {
        define('DB_USERNAME','root');
        define('DB_PASSWORD','');
        define('DB_NAME','demo1');
        define('DB_HOST','localhost');
        $con = new mysqli(DB_HOST,DB_USERNAME,DB_PASSWORD,DB_NAME);

        if (mysqli_connect_errno())
        {
            echo "Failed to connect to MySQL: " . mysqli_connect_error();
        }
        echo "Connection success...";
        return $this->con;
    } 
}

Answer 1

您的问题是在$con

中设置connect()

您正在使用

// $con is only visible within the scope of the connect function
$con = new mysqli(DB_HOST,DB_USERNAME,DB_PASSWORD,DB_NAME);

但它应该是

// this will set the config class's $con property so that it is re-usable
$this->con = new mysqli(DB_HOST,DB_USERNAME,DB_PASSWORD,DB_NAME);

$ this-＆gt; con＆gt;准备不工作

1 个答案: