Question

我正在使用php处理基本的用户操作系统，但是我无法链接dbconnect并从中获取连接变量。

这是我的代码：

dbconnect.php

<?php
$con = mysqli_connect("localhost","root","","realestatecompany");
if(!$con) {
 echo "Error";
}
//$con->close();
?>

objects.php

<?php 
require_once 'dbconnector.php';
class userAccount {
    public $userName;
    public $userPass;
    public $userType;
      public function __construct ($UserName, $Password, $Type, $con) {
        $this->userName = $UserName;
        $this->userPass = $Password;
        $this->userType = $Type;
    }
    public function writeInDB() {
        $sql = "INSERT INTO login (userName,userPass,userType) values (".$this->userName.",".$this->userPass.",".$this->userType.")";
        if ($con->query($sql) === TRUE) {
            echo "Record Added successfully";
        } else {
            echo "Error Adding record: " . $conn->error;
        }
    }
    public function editInDB($userNewAccount, $userID) {
        $a = $userNewAccount->userName;
        $b = $userNewAccount->userPass;
        $c = $userNewAccount->userType;
        $sql = "UPDATE login SET userName =".$a.", userPass ='".$b."', userType =".$c." WHERE userID =".$userID;
        echo $sql;
    }
}
$x = new userAccount("Heba", "Taba3", 1, $con);
$x->writeInDB();

＆GT;

输出

 Notice: Undefined variable: con in D:\xampp\htdocs\swproject\objects.php on line 14

 Fatal error: Uncaught Error: Call to a member function query() on unknown in D:\xampp\htdocs\swproject\objects.php:14 Stack trace: #0 D:\xampp\htdocs\swproject\objects.php(29): userAccount->writeInDB() #1 {main} thrown in D:\xampp\htdocs\swproject\objects.php on line 14

任何帮助？

Answer 1

您的问题是可变范围，而且您忘记将$con参数保存为__contruct()作为您班级的财产，并且未将其用作您班级的属性

<?php 

require_once 'dbconnector.php';  //<-- should this be 'dbconnect.php`

class userAccount {
    public $userName;
    public $userPass;
    public $userType;

    public $con;

    public function __construct ($UserName, $Password, $Type, $con) {
        $this->userName = $UserName;
        $this->userPass = $Password;
        $this->userType = $Type;
        $this->con      = $con;
    }

    public function writeInDB() {
        $sql = "INSERT INTO login 
                     (userName,userPass,userType) 
                values (".$this->userName.",".$this->userPass.
                        ",".$this->userType.")";

        if ($this->con->query($sql) === TRUE) {
            echo "Record Added successfully";
        } else {
            echo "Error Adding record: " . $conn->error;
        }
    }

    public function editInDB($userNewAccount, $userID) {
        $a = $userNewAccount->userName;
        $b = $userNewAccount->userPass;
        $c = $userNewAccount->userType;
        $sql = "UPDATE login SET userName =".$a.", userPass ='".$b."', userType =".$c." WHERE userID =".$userID;
        echo $sql;
    }
}
$x = new userAccount("Heba", "Taba3", 1, $con);
$x->writeInDB();

无法包含dbconenct.php

1 个答案: