Question

Goodday，

我收到一个奇怪的错误。

这是我的代码：

<?php
$con=mysqli_connect("localhost","root","","testdatabase");

class database{
    public function select($tableName){
        $result = mysqli_query($con,"SELECT * FROM ".$tableName);       
    }       
}

$database = new database();
?>

错误消息是$ con是一个未定义的变量。但我在第2行定义$ con？

当我有var_dump con时它也会说NULL。

我做错了什么？

Answer 1

如果要访问方法或函数中的$ con var，则必须在代码中对其进行全局化：

public function select($tableName){
    global $con;
    $result = mysqli_query($con,"SELECT * FROM ".$tableName);       
}

但是你不应该那样做，全局性是邪恶的！

这就是您的代码应该如何：

<?php
class database{
    protected $con;

    public function __construct($host, $user, $password, $dbname){
        $this->con = mysqli_connect($host, $user, $password, $dbname);
    }

    public function select($tableName){
        $result = mysqli_query($this->con,"SELECT * FROM ".$tableName);       
    }       
}

$database = new database("localhost", "root", "", "testdatabase");
?>

Answer 2

连接$con在全局范围内可用。您可能想要做的是在Database对象的构造函数中注入mysql连接（以mysqli对象的形式）。这是一个简单的例子：

<?php
$mysqli= new mysqli("localhost","root","","testdatabase");

class database{
    protected $mysqli = null;
    public function __contruct(MySQLI $mysqli = NULL) {
        $this->mysqli = $mysqli;
    }

    public function select($tableName){
        $result = $this->mysqli->query("SELECT * FROM ".$tableName);       
    }       
}

$database = new database($mysqli);
?>

请注意，既然您正在尝试使用面向对象的范例，那么您也可以以面向对象的方式使用mysqli。

Answer 3

您不能在类中使用全局变量。

<?php

    class database {

        private $con;
        public $result;

        public function select($tableName){
            $this->con = mysqli_connect("localhost","root","","testdatabase");
            $this->result = mysqli_query($this->con,"SELECT * FROM ".$tableName);       
        }       
    }

    $database = new database();

    // do something with result
    if($database->result) {
       echo "Query returned something!"; 
    }
?>

Answer 4

在你的类中添加一个构造函数，并在这个函数中声明你的连接。

class database{
    private $con;
    public function __construct(){
       $this->con=mysqli_connect("localhost","root","","testdatabase"); 
    }
    public function select($tableName){
        $result = mysqli_query($this->con,"SELECT * FROM ".$tableName);       
    }       
}

$database = new database();

php mysqli数据库连接失败？

4 个答案: