Reputation: 133

guestbook ajax php jquery

Have been trying to make a guestbook with jquery, ajax and php, I have been able to reed and print out everything inside the database but for some reason I cant save what I write in the database and then print it out as a post in the guestbook, if someone could see what I am doing wrong I would appreciate it! (for now I only try to get the username inside the database)

this is the jquery:

$("#newPost").bind('click', function(){
    var userName = $('#userName').val();
    var message = $('#message').val();
    $.ajax({
        url: "server.php?action=newPost",
        type: "POST",
        data: {userName: userName},
        success: function(data){
            if(data == "true"){
                alert(data);
                $('#posts').prepend('<td>'+userName+'</td>');
                $('#userName').val('');
            }
            else{
                alert('Something went wrong while trying to save!');
            }
        },
        error: function(xhr, error){
            alert('Could not connect to server!');          
        }

        });
    });

this is the server.php file:

$db = mysqli_connect('localhost', 'username', 'password', 'my_database');

if(isset($GET_['action']) && $GET_['action'] == 'newPost'){
    $userName = mysqli_real_escape_string($db, POST_['userName']);
    if(mysqli_query($db, "INSERT INTO message (name) VALUES ('$userName')")){
        echo "true";
    }
    else{
        echo "false";
    }

}

and tis is the html form:

<form action="#"> 
    <p>Name:</p>
    <textarea type="text" class="field" id="userName" rows="1" cols="20"></textarea><br/><br/>
    <p>Meddelande:</p>
    <textarea type="text" class="field" id="message" rows="3" cols="20"></textarea><br/><br/>
    <input value="Send" class="button" type="button" id="newPost"></input><br/>
</form>

Upvotes: 0