2016-12-08 47 views
0

我創建這需要註冊和登錄頁面的網頁,但我得到了以下錯誤:警告:mysqli_query()預計至少2個參數,1給出... register.php線64

警告:mysqli_query()預計至少2個參數,1 C中給出:\深淵Web服務器\ htdocs中\上線項目\ register.php 64

這裏是我的register.php

<?php 
ob_start(); 
session_start(); 
if(isset($_SESSION['user'])!=""){ 
    header("Location: home.php"); 
} 
include_once 'connect_db.php'; 

$error = false; 

if (isset($_POST['btn-signup'])) { 

    // clean user inputs to prevent sql injections 
    $name = trim($_POST['name']); 
    $name = strip_tags($name); 
    $name = htmlspecialchars($name); 

    $email = trim($_POST['email']); 
    $email = strip_tags($email); 
    $email = htmlspecialchars($email); 

    $pass = trim($_POST['pass']); 
    $pass = strip_tags($pass); 
    $pass = htmlspecialchars($pass); 

    // basic name validation 
    if (empty($name)) { 
     $error = true; 
     $nameError = "Please enter your full name."; 
    } else if (strlen($name) < 3) { 
     $error = true; 
     $nameError = "Name must have atleast 3 characters."; 
    } else if (!preg_match("/^[a-zA-Z ]+$/",$name)) { 
     $error = true; 
     $nameError = "Name must contain alphabets and space."; 
    } 

    //basic email validation 
    if (!filter_var($email,FILTER_VALIDATE_EMAIL)) { 
     $error = true; 
     $emailError = "Please enter valid email address."; 
    } else { 
     // check email exist or not 
     $query = "SELECT userEmail FROM users WHERE userEmail='$email'"; 
     $result = mysqli_query($dbc, $query); 
     $count = mysqli_num_rows($result); 

     if($count!=0){ 
      $error = true; 
      $emailError = "Provided Email is already in use."; 
     } 
    } 
    // password validation 
    if (empty($pass)){ 
     $error = true; 
     $passError = "Please enter password."; 
    } else if(strlen($pass) < 6) { 
     $error = true; 
     $passError = "Password must have atleast 6 characters."; 
    } 

    // password encrypt using SHA256(); 
    $password = hash('sha256', $pass); 

    // if there's no error, continue to signup 
    if(!$error) { 

     $query = "INSERT INTO users(userName,userEmail,userPass) VALUES('$name','$email','$password')"; 
     $res = mysqli_query($query); 

     if ($res) { 
      $errTyp = "success"; 
      $errMSG = "Successfully registered, you may login now"; 
      unset($name); 
      unset($email); 
      unset($pass); 
     } else { 
      $errTyp = "danger"; 
      $errMSG = "Something went wrong, try again later..."; 
     } 

    } 


} 

代碼?>

mysqli_num_rows()函數返回結果集中的行數。 因此,如果沒有行返回或行不正確,則返回FALSE。 如果mysqli_query返回一個布爾值,那麼只有在查詢失敗時纔會返回錯誤。

我不太確定如何去糾正這個錯誤,並且需要一些指導,謝謝。

+0

** WARNING **:當使用'mysqli'你應該使用[參數化查詢(HTTP:// PHP。 net/manual/en/mysqli.quickstart.prepared-statements.php)和['bind_param'](http://php.net/manual/en/mysqli-stmt.bind-param.php)添加用戶數據到您的查詢。 **不要**使用手動轉義和字符串插值或串聯來實現此目的,因爲如果您忘記正確地轉義某些內容,您將創建嚴重的[SQL注入漏洞](http://bobby-tables.com/)。 'htmlspecialchars'是**沒有辦法**一個適當的轉義SQL的方法,並會破壞否則有效的數據。 – tadman

+0

**警告**:編寫您自己的訪問控制層並不容易,並且有很多機會使其嚴重錯誤。請不要在[Laravel](http://laravel.com/)等任何現代開發框架(http://codegeekz.com/best-php-frameworks-for-developers/)上編寫自己的認證系統,內置了強大的[認證系統](https://laravel.com/docs/5.2/authentication)。絕對不會遵循[推薦的安全最佳實踐](http://www.phptherightway.com/#security)和**從不存儲具有弱無限散列**的密碼。 – tadman

回答

0

您需要連接句柄傳遞到您的mysqli_query()在線64:

$res = mysqli_query($dbc, $query); 
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