传递值到类

Question

<?php 
class RegisterUser{ 
    var $email; 
    var $password; 
    var $password_hash; 
    var $roleid; 
    public function __construct($email, $password, $roleid){ 
    $this->email = $email; 
    $this->password = $password; 
    $this->roleid = $roleid; 
    $this->addUsertoDB(); 
    } 

    public function addUsertoDB(){ 
    $this->password_hash = md5($this->password); 
    $checkemail = mysql_query("SELECT * FROM users WHERE (email ='" . mysql_real_escape_string($this->email) . "')"); 
    if(mysql_num_rows($checkemail) == 1){ 
     echo '<script type="text/javascript"> alert ("Email already used!");</script>'; 
    } 
    else{ 
     $register = "INSERT INTO users(email, password, roleid) VALUES('$this->email', '$this->password_hash', '$this->$roleid')"; 
     if($query_run = mysql_query($register)){ 
     echo('<scrip type="text/javascript"> alert("Registration complete!"); location.replace("login")</script>'); 
     } 
     else{ 
     die('<script type="text/javascript"> alert("Error inserting data!");</script>'); 
     } 
    } 
    } 
} 

$connect = new RegisterUser('carloadap@htomail.com', 123, 1); 
?> 

这是我的代码，我不断得到一个错误，我不知道如何弄清楚。我希望你们做到了。

这是错误。

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\myPHPWebsite\include\classRegister.php on line 17

Catchable fatal error: Object of class RegisterUser could not be converted to string in C:\xampp\htdocs\myPHPWebsite\include\classRegister.php on line 21

更新

<?php 
require 'connectDB.php'; 
class RegisterUser{ 
    var $email; 
    var $password; 
    var $password_hash; 
    var $roleid; 
    public function __construct($email, $password, $roleid){ 
    $this->email = $email; 
    $this->password = $password; 
    $this->roleid = $roleid; 
    $this->addUsertoDB(); 
    } 

    public function addUsertoDB(){ 
    $this->password_hash = md5($this->password); 
    $checkemail = mysql_query("SELECT * FROM users WHERE (email ='" . mysql_real_escape_string($this->email) . "')"); 
    if(mysql_num_rows($checkemail) == 1){ 
     echo '<script type="text/javascript"> alert ("Email already used!");</script>'; 
    } 
    else{ 
     $register = "INSERT INTO users(email, password, roleid) VALUES('$this->email', '$this->password_hash', '$this->$roleid')"; 
     if($query_run = mysql_query($register)){ 
     echo('<scrip type="text/javascript"> alert("Registration complete!"); location.replace("login")</script>'); 
     } 
     else{ 
     die('<script type="text/javascript"> alert("Error inserting data!");</script>'); 
     } 
    } 
    } 
} 

$connect = new RegisterUser('carloadap@htomail.com', 123, 1); 
?> 

这是错误我得到

Connected to Database! Catchable fatal error: Object of class RegisterUser could not be converted to string in C:\xampp\htdocs\myPHPWebsite\include\classRegister.php on line 22

Answer 1

第一个错误是由什么是错在你的MySQL。尝试在查询后添加
if($e = mysql_error()) die("MySQL error: ".$e);。

另外，您的可捕捉的致命错误是由$this->roleid错误地写为$this->$roleid造成的。另外请注意，您无法在该行上转义字符串，因此您很容易被注入。这是不够的，只有确保第一次查询时，你有充分的时间去做;）

Answer 2

形式，你给我们

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in 
C:\xampp\htdocs\myPHPWebsite\include\classRegister.php on line 17 

好像$checkmail值无效mysql_num_rows()结果的误差。检查你的查询，并尝试在你的数据库控制台上运行它（如果你使用的话，可以使用phpmyadmin）。

public function addUsertoDB(){ 
    ... 
    $query = "SELECT * FROM users WHERE (email ='" . mysql_real_escape_string($this->email) . "')"; 
    checkemail = mysql_query($query) or die(mysql_errno() . " : " . mysql_error() . " : " . $query); 
    ...