PHP-从另一个类调用数据库类

Question

我最近在代码中遇到错误：

PHP Parse error:  syntax error, unexpected '$connection' (T_VARIABLE), expecting function (T_FUNCTION) in

从我搜索过的问题来看，问题不在于之前未设置公共/私有/受保护，还是没有在构造函数中创建数据库连接。

数据库连接类代码：

 class databaseConnection {


    //Database information
    protected $mysqliUser = "";
    protected $mysqliHost = "localhost";
    protected $mysqlipass = "";
    protected $mysqlidbname = "";
    public $con;


    public function __construct() {
        $this->con = new mysqli($this->mysqliHost,$this->mysqliUser,$this-
        >mysqlipass,$this->mysqlidbname); 

    }

}

收到错误的类代码片段：

private $conn;
    $conn = new databaseConnection;

非常感谢您为我指明正确的方向。

Answer 1

私有声明上方只是类创建类createSession

在这种情况下， $conn = new databaseConnection; 需要可见性，或者应该在方法中。 在不查看所有类和奇数缩进的情况下很难说出哪个。

因此，将其放在方法所属的方法中（也许是构造函数？），或将其设置为私有，受保护或公共（无论打算使用哪种方法）。

Answer 2

看一下这个。

真的行..

connect.php

class DBConnect {
    private static $connection;
    private static $host = "localhost";
    private static $user = "root";
    private static $pwd = "";
    private static $dbname = "yourDBName";

    public static function connect() {
        $host = self::$host;
        $user = self::$user;
        $pwd = self::$pwd;
        $dbname = self::$dbname;

        self::$connection = new mysqli($host, $user, $pwd, $dbname) or die('Error connecting..');
        return self::$connection;
    }
}

myclass.php

require_once('connect.php');

class MyClass {
    private $connObj;

    public function __construct() {     
        $this->connObj = DBConnect::connect();
    }

    public function select() {
        $conn = $this->connObj;
        $sql = "SELECT * FROM table_name";

        $query = $conn->query($sql);
        $res = $query->fetch_assoc();
        // your code..
    }
}

$obj = new MyClass();
$obj->select();

希望，此代码段对您有效。 :) :)