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

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);
?>

What error do you get?

DevAlien
– DevAlien

2015-09-18 06:58:22 +00:00
Commented Sep 18, 2015 at 6:58 — DevAlien
– DevAlien, Commented Sep 18, 2015 at 6:58
@DevAlien 500 Internal Server Error Thanks!

Sjay
– Sjay

2015-09-18 07:00:24 +00:00
Commented Sep 18, 2015 at 7:00 — Sjay
– Sjay, Commented Sep 18, 2015 at 7:00

Norlihazmey Ghazali · Accepted Answer · 2015-09-18 07:00:26Z

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.

answered Sep 18, 2015 at 7:00

Norlihazmey Ghazali

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