在php中的注册表格

Question

$error1=''; 
$error2=''; 
$error3=''; 
$error4=''; 
$error5=''; 
$error6=''; 

$yourname=''; 
$email=''; 
$email2=''; 
$password=''; 
$password2=''; 
$country=''; 

    if (isset($_POST['Registerme'])) 
     { 

$_POST['yourname']=$yourname; 
$_POST['email']=$email; 
$_POST['email2']=$email2; 
$_POST['password']=$password; 
$_POST['password2']=$password2; 
$_POST['country']=$country; 

if($yourname==''){ 

    $error1='name required'; 

    } 


if($email==''){ 

    $error2='email required'; 

    } 
if($email2==''){ 

    $error3='required field'; 

    } 


if($password==''){ 

    $error4='password required'; 

    } 


    if($password2==''){ 

    $error5='required field'; 

    } 


if($country==''){ 

    $error6='country required'; 

    } 



     if(empty($error1) && empty($error2) && empty($error3) && empty($error4) &&  empty($error5) && empty($error6)) 

     {echo 'mysql query goes here and add the user to database';} 

     }///main one 

     else {$error1=''; 
    $error2=''; 
     $error3=''; 
     $error4=''; 
      $error5=''; 
      $error6='';} 

这是一个注册验证脚本。在我的注册表单中有两个电子邮件和密码filelds.second字段用于确认。我想检查天气用户在这两个字段中键入相同的信息。如果我想在这个脚本中这样做，我应该使用另一个if语句吗？或者我应该使用其他如果？我对这一步感到困惑......

Answer 1

一些意见：

• 您必须清理输入！看看best method for sanitizing user input with php。
• 你的任务：取代“$ _POST ['yourname'] = $ yourname;”它应该是“$ yourname = $ _ POST ['yourname'];”。
• 您正在使用很多变量进行错误控制，之后如果一切顺利，您只需忘记最后一个else块中的错误消息。使用某种数组作为错误字符串，并使用它！
• 你确定你没有验证用户名/密码不包含空格或怪异字符，或电子邮件有效吗？

一些示例代码...：

// Simple sanitize function, complete it 
function sanitize_input ($inputstr) { 
    return trim(mysql_real_escape_string($inputstr)); 
} 

if (isset ($_POST['Registerme']) { 
    // array of error messages to report 
    $error_messages = array(); 
    $isvalid = true; 

    // Assignment 
    $yourname = sanitize_input ($_POST['yourname']); 
    $email = sanitize_input ($_POST['email']); 
    $email2 = sanitize_input ($_POST['email2']); 
    $password = sanitize_input ($_POST['password']); 
    $password2 = sanitize_input ($_POST['password2']); 
    $country = sanitize_input ($_POST['country']); 

    // Validation 
    if (empty ($yourname)) { 
     $error_messages[] = "You must provide an username"; 
    } 

    if (empty ($password)) { 
     $error_messages[] = "You must provide a password."; 
    } 
    elseif ($password !== $password2) { 
     $error_messages[] = "Passwords do not match."; 
    } 

    // Same for email, you caught the idea 

    // Finally, execute mysql code if all ok 
    if (empty($error_messages)) { 
     // Execute mysql code 
     isvalid = true; 
    } 
} 

// After form processing, use isvalid which is false if there are errors 
// and the error_messages array to report errors 

Answer 2

为你的第二条if语句添加附加条件。 例如

if($email=='' || $email != $email2){ 
... 

Answer 3

只需添加简单的检查。我不会把检查与一般密码检查结合起来 - 因为我可以想象你想告诉用户出了什么问题刚好是。

if ($password1 !== $password2) { 
    // Add an specific error saying the passwords do not match. 
} 

我将取代宽松错误的用户喜欢的数组：

$aErrors = array(); 

if ($password1 !== $password2) { 
    $aErrors[] = 'Another specific error!'; 
} 

if (empty($password1) || empty($password2)) { 
    $aErrors[] = 'Another specific error'; 
} 

if (empty($aErrors)) { 
    // Process the form! 
} 

Answer 4

有许多与你的代码问题。

1. You are assinging $_POST['key'] = $somevalue, while I think you mean $somevar = $_POST['key'] 
2. Use an array for all error messages as it'll make your life a bit easier .. 
3. To compare password use something like 
if ($password1 !== $password2) { 
} 

所以.....

$errors = array(); 

所以你最好检查类似的东西..

if ($password1 !== $password2) { 
    $errors[] = 'Password dont match'; 
} 

if(count($errors) > 0) { //if there are errors 
foreach($errors as $err) { 
    echo $err.' <br />'; 
} 
} else { 
// whatever you want to do if no error 
} 

，我也会推荐您使用前消毒$ _ POST值他们在你的查询。 我希望它有帮助。

Answer 5

我想你的意思要做到这一点：

$yourname = $_POST['yourname']; 
$email = $_POST['email']; 
$email2 = $_POST['email2']; 
$password = $_POST['password']; 
$password2 = $_POST['password2']; 
$country = $_POST['country']; 

二本作使用错误数组：

$errors = array(); 

三使用嵌套IFS（只是一个建议）

 if (!empty($_POST['password1'])) { 
      if ($_POST['password1'] != $_POST['password2']) { 
       $errors[] = '<font color="red">The 2 passwords you have entered do not match.</font>'; 
      } else { 
       $password = $_POST['password1']; 
      } 
     } else { 
      $errors[] = '<font color="red">Please provide a password.</font>'; 
     }