Foreach is does not working in PHP?

Question

My PHP class:

class myclass
{
    function connection()
    {
        $con = mysqli_connect("localhost", "root", "", "clinic management system");
        return $con;
    }
    function getusername()
    {
        $q = mysqli_query($this->connection(), "select * from clients");
        return $q;
    }
}

php code:

<?php
$obj = new myclass();
$user = $obj->getalluser();
$username = array();
foreach($user as $suser){
    $username[] = $suser['client_name'];

    $data = array( "username" => "");
    if( isset($_POST["naam"]) ) {
        if( in_array( $_POST["naam"], $username ) ) {
            $data["username"] = "inuse";
        }
    }
}

echo json_encode( $data );
?>

jquery:

$(document).ready(function() {
    $("form.register").change(function() {
        $.post("check.php", $("form.register").serialize(), function(data) {
            if (data.username == "inuse")
                $("p#username_error").slideDown();
            else
                $("p#username_error").hide();
        }, "json");
    });
});

html:

<div id="myModal2" class="modal fade" role="dialog">
    <div class="modal-dialog">
        <!-- Modal content-->
        <div class="modal-content">
           <div class="modal-header">
              <button type="button" class="close" data-dismiss="modal">&times;</button>
              <h4 class="modal-title">Registeration</h4>
           </div>
           <div class="modal-body">
              <form method="post" action="index.php" name="myform" id="myform" class="register">
                 <label class="">Username:</label>
                 <input type="text" name="naam" class="form-control">
                 <p id="username_error" style="display: none;" class="error">That Username is unavailable</p>
                 <label class="">Age:</label>
                 <input type="text"  name="age" class="form-control">
                 <label class="">Gender:</label>
                 <select name="gender" class="form-control">
                    <option value="0">select gender</option>
                    <option>male</option>
                    <option>female</option>
                 </select>
                 <label class="">Email:</label>
                 <input type="text" name="email" class="form-control">
                 <label class="">phone:</label>
                 <input type="text" name="phone" class="form-control">
                 <label class="">password:</label>
                 <input type="password" name="pass" class="form-control">
                 <label class="">city:</label>
                 <input type="text" name="city" class="form-control">
                 <label class="">Address:</label>
                 <input type="text" name="address" class="form-control">
                 <input type="hidden" name="status" value="1" class="form-control">
                 <br>
                 <button class="btn btn-primary" type="submit" name="reg">register</button>
              </form>
           </div>
           <div class="modal-footer">
              <button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
           </div>
        </div>

When user type his name he get in paragraph that this user is not available if that be already exist in my database so far I have wrote this code since I am new to PHP have no idea where I am doing wrong. Why is my jQuery code not working? I have briefly define all my PHP code I written for that process also my jQuery with HTML code which I used in modal.

Answer 1

Try to update your php code like this:

PHP Code

<?php
//Include class file first
include("myclass.php");

$obj = new myclass();
$user = $obj->getusername();
$username = array();
foreach($user as $suser){
    $username[] = $suser['client_name'];
}
$data = array();
if( isset($_POST["naam"])) {
    if(in_array($_POST["naam"], $username)) {
        $data["username"] = "inuse";
    }
    else{
        $data["username"] = "not in use";
    }
}
echo json_encode($data);
?>

jQuery Code:

//callback handler for form submit
$("#myform").on('submit', function(e){
    e.preventDefault();
    var postData = $(this).serializeArray();
    var formURL = 'check.php'
    $.ajax(
    {
        url : formURL,
        type: "POST",
        dataType : "json",
        data : postData,
        success:function(data, textStatus, jqXHR) 
        {
             console.log(data);
             if (data.username == "inuse"){
                 $("p#username_error").slideDown();
             }
             else{
                 $("p#username_error").hide();
             }     
        },
        error: function(jqXHR, textStatus, errorThrown) 
        {
            console.log("Error found"); 
        }
    });
});

Answer 2

PHP Solved:

<?php
include("admin/function/kamran.php");
$obj = new myclass();
$user = $obj->getuser($_POST['a']); 

if(mysqli_num_rows($user) > 0){
echo 'inuse';
}
else{
echo 'Valid User Name';

 }

I forgot where clouse for username to get on type.

JQUERY Changes:

     $(document).ready(function(){
$("#txt").blur(function() {

    $.post("check.php", { a : $("#txt").val() } , function( data ) {
        alert(data);    
         });

     });
});

Instead sending data serialized I tried to send it with array. Now it work perfectly. Thanks for everyone contribution. Answered my own question if this code can help anyone else. html is same as posted in main post.