Question

我在PHP文件中收到此错误。

我的登录功能

<?php
session_start();
include ("../dbConnection.php");


class login {

    public $link;

    function __construct()
    {
        $dbc = new dbConnection();
        $this->link = $dbc->Connect();
        return $this->link;
    }

    public function get_data($emailid,$password)
    {
        $q = $this->link->prepare("SELECT id from students WHERE emailid='$emailid' AND password='$password' AND active='Yes'");
        $q->execute(array(':emailid'=>$emailid,':password'=>$password));
        $counts = $q->fetch();
        if($counts['id'] > 0)
        {
            session_start();
            $_SESSION['userlogin'] = $counts['id'];
            $encrypt_id1 = $this->encrypt_decrypt('encrypt', $counts['id']);
            echo $encrypt_id1;
        }            
    }
}

dbConnection.php

<?php

public class dbConnection {

    public $conn;


    public $db_host = 'localhost';
    public $db_name = 'pte_mock';
    public $db_user = 'root';
    public $db_pass = '';

    public function Connect()
    {
        try{
            $this->conn = new PDO("mysql:host=".$this->db_host.";dbname=".$this->db_name,$this->db_user,$this->db_pass);
            $this->conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
        }
        catch(PDOException $e)
        {
            echo 'ERROR: ' . $e->getMessage();
        }    
        return $this->conn;
    }
}

我收到以下声明

的错误

"$q = $this->link->prepare("SELECT id from students WHERE emailid='$emailid' AND password='$password' AND active='Yes'");"

错误

"PHP Fatal error:  Uncaught Error: Call to a member function prepare() on null"

我不知道这个问题是什么。如果有人知道解决方案，请帮我解决这个问题。提前谢谢。

Answer 1

定义类时，需要构造函数才能初始化实例变量。

class dbConnection {

    public $conn,

           $db_host,
           $db_name,
           $db_user,
           $db_pass;

    public function __construct()
    {
        $this->conn = false;

        $this->db_host = 'localhost';
        $this->db_name = 'pte_mock';
        $this->db_user = 'root';
        $this->db_pass = '';
    }

    /* ... */

也无法定义类public，这会在编译时引发语法错误。

所以你只有class dbConnection {

我不确定这是否会导致问题，但是负责人不能返回值。实际上，它们返回刚创建的实例（对象）。

你已进入班级login：

function __construct()
{
    $dbc = new dbConnection();
    $this->link = $dbc->Connect();
    // return $this->link;  // <--- REMOVE THAT - YOU CANNOT RETURN VALUES HERE!
}

当您致电$lgn = new login()时，系统会调用__construct()函数，并进入$lgn类login的新实例。如果构造函数返回任何被丢弃的内容！

所以你应该这样重构你的代码：

$lgn = new login(); // <--- returns a new instance of the class login
$the_link= $lgn->link; // <--- this way you access `link` instance variable

最后这个

    $q = $this->link->prepare("SELECT id from students WHERE emailid='$emailid' AND password='$password' AND active='Yes'");

在将值注入查询而不是指定占位符时，不是构建预准备语句的正确方法。

该行应以这种方式编写

    $q = $this->link->prepare("SELECT id from students WHERE emailid=:$emailid AND password=:password AND active='Yes'");

Answer 2

替换你的代码的这一部分

$q = $this->link->prepare("SELECT id from students WHERE emailid='$emailid' AND password='$password' AND active='Yes'");

$q->execute(array(':emailid'=>$emailid,':password'=>$password));

用这个：

$q = $this->link->prepare("SELECT id FROM students WHERE emailid = :emailid AND password = :password AND active = 'Yes'");

$q->execute(array(':emailid'=>$emailid,':password'=>$password));

PHP致命错误：未捕获错误：在null上调用成员函数prepare（）

2 个答案: