2015-01-02 28 views
0

我希望能夠禁止我的表單提交時提交用戶名,我想得到一些幫助。禁用按鈕當用戶名被採取時jquery

我的js代碼:

$('#username').keyup(function() { 
var username = $(this).val(); 

$('#username_status').html('Searching...'); 

if (username !='') { 
    $.post('username_check.php', { username: username}, function(data) { 
     $('#username_status').html(data); 
    }); 
} else{ 
    $('#username_status').html(''); 
} 
}); 

,我的檢查用戶名存在與否:

PHP頁面
<?php 
if(isset($_POST["username"])) 
{ 
define("HOST","localhost"); 
define("USERNAME","root"); 
define("PASS",""); 
define("DBNAME","testingproject"); 

$connecDB = mysqli_connect(HOST, USERNAME, PASS, DBNAME)or die('could not connect to database'); 


$username = $_POST["username"]; 


$results = mysqli_query($connecDB,"SELECT id FROM attendant WHERE username='$username'"); 

$username_exist = mysqli_num_rows($results); //records count 


if($username_exist) { 
    $output= "Sorry, this Username is taken"; 
    echo ($output); 
}else{ 
    echo('available'); 
} 
} 
?> 
+0

你可以顯示控制make keyup的代碼嗎? –

+0

將你的表單提交到一個Javascript函數中,然後你可以在檢查你的條件後使用submit()。 – Seano666

回答

0

if($username_exist) { echo 'yes'; } 
else { echo 'no'; } 

的JS,

$.post('username_check.php', { username: username}, function(data) { 
    if (data == 'yes') { 
     // here disable the registration button 
    } else { 
     // active the registration button 
    } 
}); 

希望對您有所幫助!

你也可以檢查這個線程:JQuery/Ajax Post returned data and IF function