使用PHP CodeIgniter和JQuery按下“提交”按钮之前，请检查用户名是否存在

Question

I am trying to check if a username exists even before hitting the submit button using CodeIgniter. Here is the Javascript in my view.

<script type="text/javascript" src="../assets/js/jquery.min.js"></script>
<script type="text/javascript"><!--\
$(document).ready(function(){
$('#username').keyup(username_check);
});

function username_check(){  
var username = $('#username').val();
if(username == "" || username.length < 4){
$('#username').css('border', '3px #CCC solid');
$('#tick').hide();
}else{

jQuery.ajax({
   type: "POST",
   url: "addpatient/insert",
   data: 'username='+ username,
   cache: false,
   success: function(response){
if(response == 1){
    $('#username').css('border', '3px #C33 solid'); 
    $('#tick').hide();
    $('#cross').fadeIn();
    }else{
    $('#username').css('border', '3px #090 solid');
    $('#cross').hide();
    $('#tick').fadeIn();
         }

}
});
}



}

</script>

Here is a part of the form:

<form method="post" action="addpatient/insert">
        <table id='addpatient'>
            <tr>    
                <td width=100> Username:
                <td width=150>  <input type='text' id='username' name='username'>
            </tr> .....

Here is a part of the the controller (function insert )

$patient['username'] = strtolower($this->input->post('username'));
$response = $this->user->check_username($patient['username']); 

Here is a function in the model:

function check_username_availability($username){
    $query = $this->db->query("SELECT username FROM users WHERE username = '".$username."'");
    $this->db->limit(1);
    $num = $query->num_rows();

    return $num;

 }

The result is always valid whenever I try it even though such username already exists in the database. I think my problem is in the type: "POST" of the javascript. And how could it get the number of rows fetched (fetched in the model ) to the javascript? I just pattered my code here . Please help me make this work. I'm new in using CodeIgniter framework. Thank you!

EDIT

$(document).ready(function(){
$('#username').keyup(username_check);
});

function username_check(){  
var username = $('#username').val();
if(username == "" || username.length < 4){
$('#username').css('border', '3px #CCC solid');
$('#tick').hide();
}else{

var url = 'addpatient/insert';
var data = {username:'username'};
$.post(url, data, function(result){
   console.log(result);
  function(response){
if(response == 1){
    $('#username').css('border', '3px #C33 solid'); 
    $('#tick').hide();
    $('#cross').fadeIn();
    }else{
    $('#username').css('border', '3px #090 solid');
    $('#cross').hide();
    $('#tick').fadeIn();
         }

}
}
});
}



}

Answer 1

假设您的php代码返回正确的值...在成功回调中修剪结果或返回json。

if($.trim(response) == 1)

Answer 2

Try this one

function check_username_availability($username){
    $query = $this->db->query("SELECT username FROM users WHERE username = '".$username."'");
    $this->db->limit(1);
    $num = $query->num_rows();

    echo $num; //echo the value instead return

 }

Answer 3

I think your ajax call is the problem. The data that you are posting should be an object. Try doing it in jquery. its easier:

var url = 'addpatient/insert';
var data = {username:'username'};
$.post(url, data, function(result){
   console.log(result);
   ..your success function..
});