在php类中包含文件
我第一次用php中的类做一些事情。 我正在尝试获取类中的在php类中包含文件,php,class,Php,Class,我第一次用php中的类做一些事情。 我正在尝试获取类中的返回对象数组项 这是我的课 class User { $dbconn = include("config.php"); private $dbHost = $dbconn->host; private $dbUsername = $dbconn->username; private $dbPassword = $dbconn->pass; private $dbName
返回对象数组
项
这是我的课
class User {
$dbconn = include("config.php");
private $dbHost = $dbconn->host;
private $dbUsername = $dbconn->username;
private $dbPassword = $dbconn->pass;
private $dbName = $dbconn->database;
private $userTbl = 'users';
function __construct(){
if(!isset($this->db)){
// Connect to the database
$conn = new mysqli($this->dbHost, $this->dbUsername, $this->dbPassword, $this->dbName);
if($conn->connect_error){
die("Failed to connect with MySQL: " . $conn->connect_error);
}else{
$this->db = $conn;
}
}
}
}
这是我的config.php
文件
return (object) array(
'host' => 'localhost',
'username' => 'my_user',
'pass' => 'my_pass',
'database' => 'my_db'
);
我该怎么做
PHP Parse error: syntax error, unexpected '$dbconn' (T_VARIABLE)
包括:
$dbconn = include("config.php");
在构造函数中。变量定义中不能有可执行代码,只能有静态值。所以这类事情不受支持:
class foo {
public $var = result_of_some_function();
}
如果要初始化值,请使用构造函数。您最好将其作为配置文件阅读:
class User {
public function __construct() {
$config = json_decode(file_get_contents('config.json'));
$conn = new mysqli($config->host, ...);
}
}
或者更好,使用依赖项注入:
class User {
protected $db = null;
public function __construct($db) {
$this->db = $db;
}
}
然后在创建用户对象的代码中:
$db = new Db($config);
$user = new User($db);
试试这种使用方法
<?php
class User
{
private $dbconn = null;
private $dbHost;
private $dbUsername;
private $dbPassword;
private $dbName;
private $userTbl = 'users';
function __construct()
{
include 'config.php'; //included file in constructor
if (!isset($this->db))
{
$this->dbHost= $this->dbconn->host;
$this->dbUsername= $this->dbconn->username;
$this->dbPassword= $this->dbconn->pass;
$this->dbName= $this->dbconn->database;
// Connect to the database
$conn = new mysqli($this->dbHost, $this->dbUsername, $this->dbPassword, $this->dbName);
if ($conn->connect_error)
{
die("Failed to connect with MySQL: " . $conn->connect_error);
} else
{
$this->db = $conn;
}
}
}
}
也许您应该将代码更改为
function __construct()
{
//included db file
include 'config.php';
if (!isset($this->db))
{
//code here
}
另一种方法是在配置文件中定义常量并在类中使用它们
在config.php文件中
define('HOST', 'localhost');
define('USERNAME', 'my_user');
define('PASS', 'my_pass');
define('DATABASE', 'my_db');
类内文件
include("config.php")
class User {
private $dbHost = HOST;
private $dbUsername = USERNAME;
private $dbPassword = PASS;
private $dbName = DATABASE;
private $userTbl = 'users';
function __construct(){
if(!isset($this->db)){
// Connect to the database
$conn = new mysqli($this->dbHost, $this->dbUsername, $this->dbPassword, $this->dbName);
if($conn->connect_error){
die("Failed to connect with MySQL: " . $conn->connect_error);
}else{
$this->db = $conn;
}
}
}
}
你为什么有这个$dbconn=include(“config.php”);在类之前包含该文件。@RavinderReddy我已经尝试过了。你是什么意思?我包括主机,用户名,通行证,数据库从文件。但这在课堂上不起作用。我只想感谢大家给了我各种各样的答案,这些答案对我很有帮助!谢谢这个答案我最喜欢,谢谢,我会试试的now@Darktwen欢迎尝试……:)让我知道它工作正常吗?如果在构造函数中建立连接,在对象上设置host、user、pass和dbname变量没有意义。我可以用$this->dbconn->database立即调用它,而不必在上面定义它,我明白了。可能是这样。但我所做的是以OP想要的方式回答了这个问题。
include("config.php")
class User {
private $dbHost = HOST;
private $dbUsername = USERNAME;
private $dbPassword = PASS;
private $dbName = DATABASE;
private $userTbl = 'users';
function __construct(){
if(!isset($this->db)){
// Connect to the database
$conn = new mysqli($this->dbHost, $this->dbUsername, $this->dbPassword, $this->dbName);
if($conn->connect_error){
die("Failed to connect with MySQL: " . $conn->connect_error);
}else{
$this->db = $conn;
}
}
}
}