PHP致命错误：未捕获错误：在null上调用成员函数prepare（）

我的PHP文件中出现了这个错误

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;
        }            
    }
}

错误

---

我不知道这有什么问题。如果有人知道解决方案，请帮助我解决这个问题。提前谢谢。

替换这部分代码

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

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

为此：

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

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

定义类时，确实需要构造函数来初始化实例变量

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 = '';
    }

    /* ... */

此外，类不能定义为公共类，这将在编译时引发语法错误

所以您只需要类dbConnection{

我不确定这是否会导致问题，但构造函数不能返回值。事实上，他们返回刚刚创建的实例对象

您已登录到类中：

最后这个

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

在向查询中注入值而不是指定占位符时，不是生成准备好的语句的正确方法

这行应该这样写

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

---

php中可能重复的构造函数不能返回任何东西！我尝试过，但没有解决我的问题。这不是一个准备好的语句，您只是直接在sql语句中注入变量，就像用撬棍敲打一个nail@arkascha你能支持你的评论吗？不是我的反对票。但是你指出的问题t只是其中的一个，而不是破坏代码的那个。

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 = '';
    }

    /* ... */

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

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

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

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