函数 Db::__construct() 的参数太少，传入 0
[英]Too few arguments to function Db::__construct(), 0 passed in
问题描述
i have 3 pages one for connect the DB
我有 3 页一页用于连接数据库
class Db{
private $dbUserName ="root";
private $dbName = "oop";
private $dbPas = "";
private $dbHost = "127.0.01";
public function __construct($dbUserName, $dbName, $dbPas, $dbHost)
{
$this->dbUserName = $dbUserName;
$this->dbName = $dbName;
$this->dbPas = $dbPas;
$this->dbHost = $dbHost;
$con = new PDO("mysql:host= $dbHost; $dbUserName,$dbName,$dbPas");
return $con ;
}
the 2 is 2 是
class User extends Db{
protected function getAllUser(){
$sql = "SELECT * FROM user";
$result = $this->connect()->query($sql);
$count = $result->rowCount();
// check if there'r date in the db
if($count > 0){
while($row = $result->fetchAll() ){
$data[] = $row;
}
return $data;
}
}
}
the 3 is 3 是
class ViewUser extends User{
public function ViewAllUser (){
$datas = $this->getAllUser();
foreach($datas as $data){
//echo the db rows
echo $data['uid']."</br>";
echo $data['pas']."</br>";
}
}
}
when I try to run them on another page:当我尝试在另一个页面上运行它们时：
<?php
include 'Db.php';
include 'User.php';
include 'ViewUser.php';
?>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Oop</title>
</head>
<body>
<?php
$users = new ViewUser();
$users->ViewAllUser();
?>
</body>
</html>
i get a error:我收到一个错误：
Uncaught ArgumentCountError: Too few arguments to function Db::__construct(), 0 passed in /Users/mohamedelmasry/Documents/websites/oop/test.php on line 18 and exactly 4 expected in /Users/mohamedelmasry/Documents/websites/oop/Db.php:18 Stack trace: #0 /Users/mohamedelmasry/Documents/websites/oop/test.php(18): Db->__construct() #1 /Users/mohamedelmasry/.composer/vendor/laravel/valet/server.php(128): require('/Users/mohamede...') #2 {main} thrown in /Users/mohamedelmasr
1 个解决方案
解决方案1
4 已采纳 2017-05-19 15:44:51
Your base class constructor expects 4 arguments:您的基类构造函数需要 4 个参数：
class Db{
...
public function __construct($dbUserName, $dbName, $dbPas, $dbHost)
{
...
}
So you can not create the child class without parameters.所以你不能创建没有参数的子类。 You need to either do:您需要执行以下任一操作：
$users = new ViewUser('root', 'dbname', 'pass', 'host');
Or put defaults into constructor declaration:或者将默认值放入构造函数声明中：
public function __construct($dbUserName='root', $dbName='db', $dbPas='pass', $dbHost='host')
{
...
}
But in general, the inheritance is misused here.但总的来说，继承在这里被滥用了。 You'd better have a separate Db class to manage DB and use it as a component inside your other classes (use composition instead of inheritance).您最好有一个单独的Db类来管理 DB 并将其用作其他类中的组件（使用组合而不是继承）。
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