简单的ajax代码不起作用？

<form>
<input type="text" id="user"/>
<input type="button" value="Submit" onClick="post();" />
</form>

<div id="result"> </div>

<script type="text/javascript">
   function post()
   {
       var username = $('#user').val();
       $.post('battlephp.php',
       {postuser:user}
       )       
   }
</script>

Its a simple Ajax code.. It should take username and display the Php code! But don't know why its not running?? Actually I am learning...so I cant rectify the error or fault??

I am running ii on localhost.. so is there any problem with using:

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>

display the Php code

No, it shouldn't.

First, you've changed your mind about the variable name you are using (user, username) half way through your script, so you are going to throw a reference error.

Second, you haven't provided a function (the third argument) to $.post, so you aren't doing anything (such as displaying it) with the returned data.

Third, the server should execute the PHP and return its output. You shouldn't get the actual PHP code.

function post() {
   var username = $('#user').val();
   $.post(
     'battlephp.php',
     {postuser:username}, // Be consistent about your variable names
     function (data) {
        alert(data);
     }
   );       
}

Try to do this:

<script>
    $(document).ready(function() {
        function post() {
            var username = $('#user').val();
            $.ajax({
                type : 'post',
                url : 'batttlephp.php',
                data : {
                    postuser : user
                },
                success : function(data) {
                    alert(data);
                },
                error : function(data) {
                    alert(data);
                }
            });
        });
    });
</script>

if you're doing a ajax request then is good also handle success and error... Also I suggest to you "to start the document".

Try the code above and let us know if worked

Instead

document.ready()

you can use

jQuery(function($){...});