php - Input not detected in jQuery -

this not fire code, , i'm not sure why. i've tried .change, .on('input'), can find. problem simple. wrong?

$(function() {     alert('hie');     //the min chars username       var min_chars = 3;      //result texts       var characters_error = 'minimum amount of chars 3';     var checking_html = 'checking...';      //when button clicked       $('#myusername').bind("change paste keyup", function() {         //run character number check           if ($('#myusername').val().length < min_chars) {             //if it's bellow minimum show characters_error text                              $('#username_availability_result').html(characters_error);             $('#checkuser').show(fast);             alert('hsdi');         }         else {             //else show cheking_text , run function check               $('#username_availability_result').html(checking_html);             $('#checkuser').show(fast);             check_availability();             alert('hi');         }     }); });  //function check username availability function check_availability() {      //get username       var username = $('#myusername').val();     //use ajax run check       $.post("../php/checkuser.php", {username: username},     function(result) {         //if result 1           if (result == 1) {             //show username available               $('#username_availability_result').html(username + ' available');         } else {             //show username not available               $('#username_availability_result').html(username + ' not available');         }     }); } 

html

<tr>     <td>username</td>     <td><input type="text" name="username" placeholder="username"  id="myusername"></td> </tr>   <tr id="checkuser" style="display: none">     <td>      </td> </tr> 

check sample. removed "fast" parameter .show() call.

http://jsfiddle.net/juaning/psvs7/

from calls, let me know if fix problem.

$('#checkuser').show(fast);alert('hsdi');

$('#checkuser').show(fast);