JQuery用戶名驗證

Question

我是JQuery的新手，並試圖編寫腳本來檢查用戶名可用性。我的問題是，無論我輸入什麼內容，我都會回來「此用戶名已被使用。」

的JQuery來源：

$(document).ready(function() { 

jQuery.validator.addMethod("usernameCheck", function(username) { 
    var isSuccess = false; 
    $.ajax({ url: "username_availability.php", 
      data: "username=" + username, 
      async: false, 
      success: 
       function(msg) { isSuccess = msg === "TRUE" ? true : false } 
      }); 
    return isSuccess; 
},""); 

$("#register_form").validate({ 
    onkeyup:false, 
    rules: { 
     username: { 
      required: true, 
      minlength: 3, 
      usernameCheck: true // remote check for duplicate username 
     } 
    }, 
    messages: { 
     username: { 
      required: "username is required.", 
      minlength: jQuery.format("username must be at least {0} characters in length."), 
      usernameCheck: "This username is already in use." 
     } 
    } 
}); 

PHP源：

include('database_connection.php');         
if (isset($_POST['username'])) {                
    $username = mysql_real_escape_string($_POST['username']);         
    $check_for_username = mysql_query("SELECT user_id FROM users WHERE username='$username'"); 
    if (mysql_num_rows($check_for_username)) { 
     echo "TRUE";                   
    } else { 
     echo "FALSE";                   //No Record Found - Username is available 
    } 
} 
?> 

HTML來源：

<input class="username" id="username" type="text" name="username" value="" maxlength="20" />&nbsp; 

非常感謝

Answer 1

設置Ajax調用選項type至'POST'。另外，AJAX不會阻止return isSuccess。

您可以看看jQuery Validate提供的遠程驗證。

http://docs.jquery.com/Plugins/Validation/Methods/remote#options

更改規則。

rules: { 
     username: { 
      required: true, 
      minlength: 3, 
      remote: 'username_availability.php' // remote check for duplicate username 
     } 
    }, 

請看例子。

Answer 2

我想知道如果這條線的問題是：

data: "username=" + username, 

嘗試更改爲正確的數組這是我不相信jQuery的自動解析這種格式的，尤其是對POST數據。另外，在$ .ajax調用中設置一個類型爲'post'的選項，因爲jQuery自動解析$ .ajax請求到GET，並且您在PHP腳本中使用POST。

Answer 3

嘗試改變PHP源返回 「1」 或 「0」：

include('database_connection.php');         
if (isset($_POST['username'])) {                
    $username = mysql_real_escape_string($_POST['username']);         
    $check_for_username = mysql_query("SELECT user_id FROM users WHERE username='$username'"); 
    if (mysql_num_rows($check_for_username)) { 
     echo "TRUE";                   
    } else { 
     echo "FALSE";                   //No Record Found - Username is available 
    } 
} 
?> 

，然後改變你的JS解析，作爲一個整數：

success: 
function(msg) { 
    isSuccess = parseInt(msg); 
} 

Answer 4

您可以使用下面的，只需更改它爲您的用戶名（我使用電子郵件地址作爲用戶名稱，我使用空表格單元id =「emailError」來保存該消息。

jQuery：

$('#email').keyup(function(){ 
     $.ajax({ 
     type: 'POST', 
     url: 'ajax.checkEmail.php', 
     data: {e:$('#email').val()}, 
     success: function(data){ 
      if(data == 'true'){ 
      $('#emailError').html('This email is already used'); 
      }else{ 
      $('#emailError').html(''); 
      } 
     } 
     }) 
    }); 

的PHP（ajax.checkEmail.php）：

 

    include('inc.php'); 

    $query = 'SELECT COUNT(*) FROM tblcustomers WHERE email = "'.$_POST['e'].'"'; 
    $result = $dbO->getSingleResult($query); 

    if($result > 0) 
    print 'true'; 
    else 
    print 'false'; 

Answer 5

這部作品只是檢查由相同的代碼

 jQuery.validator.addMethod("usernameCheck", function(username) { 
     var isSuccess = false; 
     $.ajax({ url: "checkusr.php", 
     data: "username=" + username, 
     async: false, 
     success: 
     function(msg) { 

     if(msg ==0) 
     { 
    isSuccess=false; 
     } 
     else 
     { 

     isSuccess=true; 
     } 
     } 


     }); 

     return isSuccess; 
     },"Username not available"); 

PHP端代碼

  if ($obj->checkUsername($_GET['username'],"artist")> 0) 
     { 
     echo 0; 
     } 
    else 
    { 
    echo 1; 
} 

    }