jQuery code not updating fields in database or updating dropdown

Question

I know that the PHP scripts work because I tested them without jQuery therefore I think something wrong with my jQuery.

<form id="bakery" name="bakery" method="POST" action="bakeryupdate.php">
            <select id="bakeryid" name="bakeryid">
                <option value="">Select</option>
                </select>
<input type="submit" value="Submit" name="submit" id="bakeryupdatebutton" />
</form>

$(document).ready(function() {

        $.ajax({
            type:'POST',
            url:'bakeryupdate.php',
        data: {bakeryid:bakeryid}
        }).done(function(){
            $('#success').text('success!');

        });
    }
        e.preventDefault();
        });    
});

update data php script

<?php       
    if(isset($_POST['submit'])) {

        $sql = "UPDATE bakeryorders SET description='shipped";

        if (mysqli_query($con, $sql)) {

            echo "Bakery order updated";
        } else {
            echo "Error!";
    }
}
?>

No errors not showing the selections in options, there should be 4 selections (i tried when i put the selection options code in the form) and when i submit it says success! but doesn't add to database and no errors for this either :( — user11119483
– user11119483, Commented Feb 11, 2015 at 17:36
mysqli_real_escape_string() requires DB connection be passed. php.net/manual/en/mysqli.real-escape-string.php - Plus, make sure your submit button has the name attribute. That conditional statement is relying on it. — Funk Forty Niner
– Funk Forty Niner, Commented Feb 11, 2015 at 17:38
Add or die(mysqli_error($con)) to mysqli_query() and use error reporting php.net/manual/en/function.error-reporting.php - You have enough to debug your code. — Funk Forty Niner
– Funk Forty Niner, Commented Feb 11, 2015 at 17:40
@Fred-ii- $bakeryid= mysqli_real_escape_string($con,$_POST['bakeryid']) ? — user11119483
– user11119483, Commented Feb 11, 2015 at 17:42

Jean-François Savard · Accepted Answer · 2015-02-11 17:36:16Z

1

Make sure to commit the transaction if you don't have autocommit...

mysqli->commit();

answered Feb 11, 2015 at 17:36

Jean-François Savard

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