Jquery检查密码与php + Json匹配

Question

我有一个表单，我正在验证JS和PHP。从我尝试检查密码是否匹配时开始，一切都进行得如此顺利。

下面的形式：

<div> 
        <label for="passlength">Password, valid: 0-9</label> 
        <input type="text" name="passlength" value="<?=@$_REQUEST['passlength']?>" id="passlength" /> 
        <span id="validatePasslength"><?php if ($error) { echo $error['msg']; } ?></span> 
       </div> 
       <div> 
        <label for="passlength">Password2, valid: 0-9</label> 
        <input type="text" name="passlength2" value="<?=@$_REQUEST['passlength2']?>" id="passlength2" /> 
        <span id="validatePasslength2"><?php if ($error) { echo $error['msg']; } ?></span> 
       </div> 

这是JavaScript：

var r = $('#passlength').val() 
; 
     var validatePasslength2 = $('#validatePasslength2'); 
     $('#passlength2').keyup(function() { 

      var t = this; 
      if (this.value != this.lastValue) { 
       if (this.timer) clearTimeout(this.timer); 
       validatePasslength2.removeClass('error').html('<img src="../../images/layout/busy.gif" height="16" width="16" /> checking availability...'); 

       this.timer = setTimeout(function() { 
        $.ajax({ 
         url: 'ajax-validation.php', 
         data: 'action=check_passlength2&passlength=' + r + '&passlength2=' + t.value, 
         dataType: 'json', 
         type: 'post', 
         success: function (j) { 
          validatePasslength2.html(j.msg); 
         } 
        }); 
       }, 200); 
       this.lastValue = this.value; 
      } 
     }); 

这里是PHP：

//Check for passlength 
     if (@$_REQUEST['action'] == 'check_passlength' && isset($_SERVER['HTTP_X_REQUESTED_WITH'])) { 
     // means it was requested via Ajax 
     echo json_encode(check_passlength($_REQUEST['passlength'])); 
     exit; // only print out the json version of the response 
    } 

    function check_passlength($password) { 
//  global $taken_usernames, $usersql; 
     $resp = array(); 
//  $password = trim($password); 
     if (!preg_match('/^[0-9]{1,30}$/', $password)) { 
     $resp = array("ok" => false, "msg" => "0-9 Only"); 
    } else if (preg_match('/^[0-9]{1,2}$/', $password)) { 
         $resp = array("ok" => false, "msg" => "Password too short"); 
     } else if (preg_match('/^[0-9]{6,30}$/', $password)) { 
           $resp = array("ok" => false, "msg" => "Password too long"); 
    } else { 
     $resp = array("ok" => true, "msg" => "Password ok"); 
    } 

     return $resp; 
    } 


//Check for passlength2 
     if (@$_REQUEST['action'] == 'check_passlength2' && isset($_SERVER['HTTP_X_REQUESTED_WITH'])) { 
     // means it was requested via Ajax 
     echo json_encode(check_passlength2($_REQUEST['passlength'],$_REQUEST['passlength2'])); 
     exit; // only print out the json version of the response 
    } 

    function check_passlength2($password,$password2) { 
//  global $taken_usernames, $usersql; 
     $resp = array(); 
//  $password = trim($password); 
     if (!preg_match('/^[0-9]{1,30}$/', $password2)) { 
       $resp = array("ok" => false, "msg" => "0-9 Only"); 
     } else if (preg_match('/^[0-9]{1,2}$/', $password2)) { 
           $resp = array("ok" => false, "msg" => "Password too short"); 
     } else if (preg_match('/^[0-9]{6,30}$/', $password2)) { 
           $resp = array("ok" => false, "msg" => "Password too long"); 
     } else if ($password !== $password2) { 
           $resp = array("ok" => false, "msg" => "Passwords do not match"); 
     } else { 
       $resp = array("ok" => true, "msg" => "Password ok"); 
     } 

     return $resp; 
    } 

我敢肯定它与问题javascript，因为如果我设置var r = 1234;有用。有任何想法吗？？

Answer 1

你只是想看看密码是否匹配，并且介于最小和最大长度之间？是不是以上的矫枉过正？我错过了什么吗？

您可以单独使用js检查第一个密码字段的长度，然后在第二个字段的onblur事件中检查field1 == field2。

我注意到次要的事情，第二个字段的标签有错误的“for”属性。