Проблем с ajax

[slideR]

Значи с един приятел направихме този код за проверка на потребителско име с ajax, но немога да разбера как трябва да е структуриран chekuser.php!

<script>
$(document).ready(function() {
$("#username").blur(function() {
$.post("styles/{AN-STYLE}/ajax/chekuser.php" , {
username : $("#username").val()
}, function(response) {
$("#userResult").fadeOut("slow");
$("#userResult").html(response);
$("#userResult").fadeIn("slow");
});
});
});
</script>

А в html слагам това

<input type="text" style="width:260px; border:1px solid #989898;" id="username" name="user" value=""><div id="userResult"></div>

Ако някой има идея плс да помага!

 

[Anonymous]

Предполагам нещо такова

<?php
include("config.php"); // или иначе казано конекцията ти към базата данни
$username = htmspecialchars(addslashes($_POST['user']));
$check = mysql_query("SELECT COUNT(`id`) FROM `table` WHERE `username` = '$username'")or die(mysql_error());
echo (mysql_result($check,0,0) > 0) ? "заето" : "свободно";
?>

 

[slideR]

Напрвих кода така

<?php
header("Content-type: text/html; charset=UTF-8");
define('IN_VALID', true);
include("../../../an_includes/connect.php");
$username = htmlspecialchars(addslashes($_POST['user']));
$check = mysql_query("SELECT COUNT(`id`) FROM `an_users` WHERE `username` = '$username'")or die(mysql_error());
echo (mysql_result($check,0,0) > 0) ? "заето" : "свободно";
?>

А това е темплейта

<div class="menu1">
<div id="text">
{MENU1}
</div>
</div>
<div class="menubg">
<form action="" method="post">
<table width="400" align="center">

<input type="hidden" name="ip" value="{IP}">
<input type="hidden" name="rank" value="1">
<tr>
<td style="border:1px solid black;">
Потребител:<font color="red">*</font>
</td>
<td style="border:1px solid black;">
<input type="text" style="width:260px; border:1px solid #989898;" id="username" name="user" value="">
<script type="text/javascript">
$(document).ready(function() {
$("#username").blur(function() {
$.post("styles/{AN-STYLE}/ajax/chekuser.php" , {
username : $("#username").val()
}, function(response) {
$("#userResult").fadeOut("slow");
$("#userResult").html(response);
$("#userResult").fadeIn("slow");
});
});
});
</script>
<div id="userResult"></div>
</td>
</tr>
<tr>
<td style="border:1px solid black;">
Парола:<font color="red">*</font>
</td>
<td style="border:1px solid black;">
<input type="password"style="width:260px; border:1px solid #989898;" name="pass" value="">
</td>
</tr>
<tr>
<td style="border:1px solid black;">
Повтори парола:<font color="red">*</font>
</td>
<td style="border:1px solid black;">
<input type="password" style="width:260px; border:1px solid #989898;" name="pass1" value="">
</td>
</tr>
<tr>
<td style="border:1px solid black;">
Email:<font color="red">*</font>
</td>
<td style="border:1px solid black;">
<input type="text" style="width:260px; border:1px solid #989898;" name="mail" value="">
</td>
</tr>
<tr>
<td style="border:1px solid black;">
Skype:
</td>
<td style="border:1px solid black;">
<input type="text" style="width:260px; border:1px solid #989898;" name="skype" value="">
</td>
</tr>
<tr>
<td style="border:1px solid black;">
Пол:
</td>
<td style="border:1px solid black;">
<select name="pol" style="width:260px; border:1px solid #989898;">
<option value="Мъж">Мъж</option>
<option value="Жена">Жена</option>
</select>
</td>
</tr>
<tr>
<td style="border:1px solid black;">
Въведи кода от картинката
</td>
<td style="border:1px solid black;">
<div align="left"><img src="an_includes/captcha.php" id="captcha"><br>
<a href="#" onclick="document.getElementById('captcha').src = 'includes/captcha.php?sid=' + Math.random(); return false">Нов код</a>
</div>

<div align="left"><input type=text name=captcha maxlength=5></div>
</td>
<tr>
<td>
<input type="submit" id="newpass" name="davai" value="Регистрирай се!">
</td>
</tr>
</table>
</form>

Винаге ми дава, че името е свободно, а в базата ина такъв потребител

 

[slideR]

Оправих се Мерси ще ти дам +1, защото по твоя код работих