ajax update to database - php

Question

I am trying to add data to a database and for some reason it does not always work. I'd say 80% of the time it will work and I'll see the result in the database but sometimes its like the script won't run.
here is the ajax :
<script type="text/javascript">
$(document).ready(function () {
$(function () {
$('.error').hide();
$("#success").hide();
$(".button").click(function () {
$('.error').hide();
var name = $("input#name").val();
if (name == "") {
$("label#name_error").show();
$("input#name").focus();
return false;
}
$.ajax({
type: 'POST',
url: "class/proccess.php",
data: $("input#name"),
cache: false,
success: function () {
$("#success").fadeIn(200).show();
}
});
});
});
});
</script>
here is the html:
<div id = "contact_form">
<form name ="contact" action="">
<fieldset>
<label for="name" id="name_label">Name</label>
<input type="text" name="name" id="name" size="30" value="" class="text-input" />
<label class="error" for="name" id="name_error">This field is required.</label>
<br />
<input type="submit" name="submit" class="button" id="submit_btn" value="send" />
</fieldset>
<span id="success"> the name has been updated successfully!</span>
</form>
</div>
<div id ="upd"></div>
and here is the proccess.php file:
<?php
$va = $_POST["name"];
$dbconnection = mysql_connect('adatabase','someuser','somepw');
if(!mysql_select_db('some_database', $dbconnection)) {
echo mysql_error();
}
else
{
echo 'connection success';
}
$sql = "INSERT INTO some_db(text) VALUES ('$va')";
$result = mysql_query($sql) or die('erreur sql!'.mysql_error());
if(!$result) {
echo "not working";
}
else {
echo "working";
}
?>
so how come it does not always insert into the database?
and is there a way to get the result from the php if(!$result) to show in the success part of the ajax?

Answer 1

You're actually passing a jQuery-Object to your PHP-File.
$.post("class/proccess.php", {
name: $("input#name").val() //Pass val() not the whole jQuery-Object!
}, function() {
/* success */
});

Answer 2

While you're debugging, make sure MySQL errors are enabled.
In your Javascript for the Ajax success handler, show an alert with the text returned from the call. That way if there's an error with MySQL you'll see it.
Another thing is, could the "text" field be set in the database as UNIQUE? Trying to insert a new record with a duplicate string would fail in that case.
And... the name of the field isn't really 'text' is it? I would recommend avoiding field names that are the same as the basic data types for MySQL. Just to avoid confusion if no other reason.

Answer 3

If it is not working sometimes, you need to check the returned string for errors. The right way to do this using AJAX is as follows.
You can include a parameter in your success callback which will fetch the page-result from the PHP.
Instead of
success: function () {
...
}
use
success: function (data) {
alert(data);
}

Answer 4

Change your ajax call to:
$.ajax({
type: 'POST',
url: "class/proccess.php",
data: {name : $("input#name").val()},
cache: false,
success: function () {
$("#success").fadeIn(200).show();
}
So that $_POST['name'] is set to the value of your input box.
Also, as suggested, you should change your mysql functions to mysqli functions to help protect against sql injections.