Update Variables in database via AJAX

Question

So, I've been practicing with PDO, see my earlier asked question and now I am stuck at the following point:

I want to update the database variables without pushing on a button, and in my opinion is it best doable via AJAX.

A few codes: General.JS

var timeoutId;
$('form input').on('input propertychange change', function() {
    console.log('Invoer bewerking');

    clearTimeout(timeoutId);
    timeoutId = setTimeout(function() {
        // Runs 1 second (1000 ms) after the last change
        saveToDB();
    }, 1000);
});

function saveToDB()
{
    console.log('Opslaan naar Database');
    form = $('.formulier24');
    $.ajax({
        url: "ajax2.php",
        type: "POST",
        data: form.serialize(), // serializes the form's elements.
        beforeSend: function(xhr) {
            // Let them know we are saving
            $('.HowAbout').html('Opslaan...');
        },
        success: function(data) { console.error(data) ;
            var jqObj = jQuery(data); // You can get data returned from your ajax call here. ex. jqObj.find('.returned-data').html()
            // Now show them we saved and when we did
            var d = new Date();
            $('.HowAbout').html('Opgeslagen om: ' + d.toLocaleTimeString());
        },
    });
}

// This is just so we don't go anywhere
// and still save if you submit the form
$('.formulier24').submit(function(e) {
    saveToDB();
    e.preventDefault();
});

ajax.php FILE:

<?php
include('verbinding.php');
if(isset($_POST['formulier24'])) {
    $sql = "UPDATE INTO evenement SET username = :username, hours = :hours";
    $parameters = array($_POST["username"], $_POST["hours"]);
    try {
        $stmt = $conn->prepare($sql);
        $stmt->bindParam(':username', $_POST['username']);
        $stmt->bindParam(':hours', $_POST['hours']);
        $result = $stmt->execute($parameters);
        $return = "Updated data successfully!";
    } catch (PDOException $e) {
        $return = "Could not update data! Error: " . $e->getMessage();
    }

    header("Content-Type: application/json");
    echo json_encode($return);
}
?>

The connection file (verbinding.php) is 100% working. I think I made a fault in the ajax.php file but I dont know where. Please let me know what I did wrong, because I do not get the Saved at time date-mention and it does not save in the database. Thanks in advance!

yes what is the error? and for the "improving code" portion of question, you should go to different stack exchange site, i think "code review" — Andrew, Mar 13 '17 at 20:09
Your update syntax might be the problem, supposed to be : Update tableName Set columnName =value where columnane = value — Masivuye Cokile, Mar 13 '17 at 20:15
Echo out the $sql that is generated by the pdo, to make sure it is what you expect. $stmt->queryString. — Andrew, Mar 13 '17 at 20:16
@MasivuyeCokile I tried, but without succes. Thanks for thinking with me tho. — Willem-Jelle, Mar 13 '17 at 20:18
@Andrew Thats this part right: `$stmt->bindParam(':username', $_POST['username']);` — Willem-Jelle, Mar 13 '17 at 20:19
oh right that doesn't exist, you need to add your own function to pull the paramaterized query. EG: http://stackoverflow.com/questions/210564/getting-raw-sql-query-string-from-pdo-prepared-statements In any case, your query looks ok, but most likelu your POSTed variables are invalid/ the name's in the form are incorrect, do var_dump($_POST) at the top to see what is going on — Andrew, Mar 13 '17 at 20:29

GertG · Answer 1 · 2017-03-14T14:30:21.927

0

Your PHP code seems to be expecting a POST variable that represents the form, but that's not what jQuery serialize produces. It produces a POST string containing the names of all the (successful) controls and their values, so PHP will receive a POST parameter per control.

EDIT: To clarify, what I mean is that your isset check on $_POST['formulier24'] will always return false if formulier24 is the name of the form.

edited Mar 14 '17 at 14:30

answered Mar 14 '17 at 11:31

GertG

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But what do I have to put over there in stead of `$_POST['formulier24']`. Also, when I put there something else than the **Saved at time/date** won't show up. – Willem-Jelle Mar 15 '17 at 12:28
I can't write your code for you. Start by looking at what you're actually sending as POST variables, e.g. by doing a `console.log` of `form.serialize()`, so that you understand what you're sending and what the PHP script can work with. – GertG Mar 15 '17 at 13:13

Update Variables in database via AJAX

1 Answers1