Insert Into MYSQL DB using jquery/ajax 500 internal server error - php

Question

I'm inserting radio checked data values into mysql db with php/jquery and ajax while inserting values i'm getting internal server 500 error...
Any help is appreciated Thanks!
Html Code
<div>
<input class="radio_div_clicked" value="<?php echo $row['name'];?> name="radio_section" id="<?php echo $row['id'];?>"/>
<p><?php echo $row['name'];?></p>
<button id="add_feed">Add</button>
</div>
JQUERY CODE
<script type="text/javascript">
$('body').delegate('#add_feed','click',function(){
var radio_div_clicked = $('input:radio[name=radio_section]:checked').val();
$.ajax({
type:"POST",
url:"index.php",
cache: false,
data:{
insert_section:1,
radio_div_clicked:radio_div_clicked
},
success:function(data)
{
alert('Successfully Inserted Into Table');
}
});
}
});
</script>
PHP code
<?php
$con = #mysql_connect('localhost','root','') or die('failed to connect server');
$db = mysql_select_db('demo',$con) or die('failed to select db');
if(isset($_POST['insert_section']))
{
$radio_div_clicked = mysql_real_escape_string($_POST['radio_div_clicked']);
$sql = mysql_query("insert into test(name) values($radio_div_clicked)";
if($sql)
{
echo "Successfully Inserted Into Table";
}
else
{
echo "Failed To Insert Value Into Table"
}
}
?>
<?php
mysql_close ($con);
?>

Answer 1

Mysql query values($radio_div_clicked)"; did't closed properly, should be
$sql = mysql_query("insert into test(name) values('".$radio_div_clicked."')");
If you're using chrome, click on network tab, and click the requested link. Then, again click on preview or response header. Usually the details errors showing up there.