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PHP array to seperate JavaScript file via AJAX

I made a simple php file, that saves data from MySQL db into 2 arrays. I am trying to send these two arrays to the js file (which is on seperate from the html file). I am trying to learn AJAX, but it seems i am not doing something correct.
Can you please explain what am i doing wrong?
My php file: get.php
<?php
define('DB_NAME', 'mouse');
define('DB_USER', 'root');
define('DB_PASS', '');
define('DB_HOST', 'localhost');
$link = mysqli_connect(DB_HOST, DB_USER, DB_PASS, DB_NAME);
if($link === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
}else{
echo 'Successfuly connected to database :) <br/>';
}
$sql = "SELECT x, y FROM mousetest";
$result = mysqli_query($link, $sql);
$x_array = [];
$y_array = [];
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
echo "x: " . $row["x"]. " - y: " . $row["y"]. "<br>";
array_push($x_array, $row["x"]);
array_push($y_array, $row["y"]);
}
} else {
echo "0 results";
}
echo json_encode($x_array);
echo "<br>";
echo json_encode($y_array);
mysqli_close($link);
$cd_answer = json_encode($x_array);
echo ($cd_answer);
?>
And this is my JS file:
$(document).ready(function(){
$.ajax({
type: "GET",
url: "get.php",
dataType: "json",
data : {anything : 1},
success:function(data){
var x = jQuery.parseJSON(data); // parse the answer
x = eval(x);
console.log(x.length);
}
});
});
I really hope you understand, what i am trying to do. Where is the problem? I really thought this should work, as i went through it quite a few times to say the least...

You can't use echo json_encode(...) twice. The client expects a single JSON object, not a series of them.
You should make each array an element of a containing array, which you then return as JSON.
$result = array('x' => $x_array, 'y' => $y_array);
echo json_encode($result);
Then in the jQuery code you would use:
var x = data.x;
var y = data.y;
Also, when you use dataType: 'json', jQuery automatically parses the JSON when it sets data. You shouldn't call jQuery.parseJSON() or eval().

parse return data from html to ajax

I have some problem with the returned value of ajax.
this is the ajax code:
$(document).ready(function() {
var request;
$("#flog").submit(function(event) {
if(request)
request.abort();
event.preventDefault();
var form = $(this);
var serializedData = form.serialize();
var btnname = $('#log').attr('name');
var btnval = $('#log').val();
var btn = '&'+btnname+'='+btnval;
serializedData += btn;
request = $.ajax({
type: form.attr('method'),
url: form.attr('action'),
data: serializedData,
});
request.done(function(data, status, jdXHR) {
alert(data);
});
request.fail(function(jdXHR, status, error) {
});
});
});
it takes data from a form and send it to another page.
this is the second page:
<?php include 'head.php'; ?>
<?php
if($_POST['login']) {
session_regenerate_id(true);
$con = mysqli_connect("localhost", "Alessandro", "ciao", "freetime")
or die('Could not connect: ' . mysqli_error($con));
$query = 'SELECT * FROM users WHERE username="' . $_POST['user'] . '"';
$result = mysqli_query($con, $query) or die('Query failed: ' . mysqli_error($con));
if(mysqli_num_rows($result) == 0) {
mysqli_close($con);
session_unset();
session_destroy();
$res = false;
return $res;
}
$query = 'SELECT password FROM users WHERE username="' . $_POST['user'] . '"';
$result = mysqli_query($con, $query) or die('Query failed: ' . mysqli_error($con));
$line = mysqli_fetch_array($result, MYSQL_ASSOC);
if(md5($_POST['password']) != $line['password']) {
mysqli_close($con);
session_unset();
session_destroy();
return false;
}
?>
<?php include 'foot.php'; ?>
and in .done the returned data is all the html page.
How can I retrieve only a data, like $res? I tried with json_encode() but with no results.
If in the second page I delete the lines include 'head.php' and include 'foot.php' it works. But I need that the secon page is html, too.
Somenone can help me?

Dont use the Data attribute from AJAX.
Replace
request.done(function(data, status, jdXHR) {
alert(data);
});
with
request.done(function(data, status, jdXHR) {
alert(jdXHR.responseText);
});

You could do it in a much simpler way.
In PHP store the result of the attempted login into a variable, for instance $result =0; to start with
If the login is valid change it to 1 and return it to ajax by doing an echo at the end of your PHP file. If you need other value returned such as the name you could add it to the variable with a separator such as || for instance.
in javascript collect your return and go data = data.split('||');
if (data[0] == 0){alert("Welcome back " + data[1]);}else{alert("wrong login...")}
Previous use is correct, you need to escape the user collected in your PHP script.
Hope this helps.

Insert into MySQL through a Buttons OnClick

Using a few answers on here I have got row being added to MySQL upon a button press but the data is blank and so I can only assume the variables are not being passed.
I really don't know what I am doing wrong, any help would be greatly appreciated.
PHP
<? $sql = "SELECT itemname FROM items ORDER BY itemname ASC";
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result)) {
echo "<button onclick='javascript:ajaxCall(" . $row['id'] . ")'><span class='btn-text'>" . $row['itemname'] . "</span></button>";
}
?>
jQuery
<script src="http://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script>
function ajaxCall(id){
$.ajax({
type: "POST",
url: "additem.php",
success: function(data){
// callback function
}
});
return false;
}
</script>
additem.php
// Connect database.
include("settings.php");
mysql_connect($db_host,$db_user,$db_pass);
mysql_select_db($db_name);
$id = $_POST['id'];
$itemsearch = mysql_query("SELECT itemname, itemcategory, price, qty FROM presales WHERE id='$id'");
$itemsearchrest = mysql_num_rows($itemsearch);
$itemname = $itemsearchrest['itemname'];
$itemcategory = $itemsearchrest['itemcategory'];
$price = $itemsearchrest['price'];
$qty = $itemsearchrest['qty'];
$sql = "INSERT INTO presales (itemname, itemcategory, price, qty) VALUES('$itemname', '$itemcategory', '$price', '0')";
if(mysql_query($sql)){
return "success!";
}
else {
return "failed!";
}
?>

mysql_num_rows returns the number of rows. It's not an array. Use fetch_assoc or similiar.
See sample in the PHP documentation!
Also your AJAX call is missing the data:
$.ajax({
type: "POST",
url: "additem.php",
data: {
id: id
}
});
Please switch to PDO or MySQLi. MySQLi will use the same function names but it is object orientated. PDO will name the functions slightly different but basically work the same way.

Using json_encode without displaying it on my webpage via echo or print

basically I'm trying to pass an array from PHP to JavaScript, so far it is all working the methods I'm using are:
PHP:
echo json_encode($arrayname);
JavaScript:
$.getJSON( 'myphppage.php', {}, function(data){
// Do stuff here
});
Obviously this echo's the text onto my webpage but I do not want this text to be displayed, I'm just wondering if there is anyway for me to use this without having a chunky array at the top of my webpage. (I tried it without the echo and it doesn't work, I've also gone through countless tutorials on this but no one seems to do it without using echo)
Thanks a lot in advance
---------- Edit -------------
index.js
$.getJSON( 'myphppage.php', {}, function(data){
// I loop through the data here
}
}).done(function() {});
myphppage.php
<?php
$servername = "name";
$username = "username";
$password = "";
$dbname = "dbname";
$connection = mysql_connect($servername,$username);
if(!$connection) {
die("Database connection failed: " . mysql_error());
}else{
$db_select = mysql_select_db($dbname,$connection);
if (!$db_select) {
die("Database selection failed:: " . mysql_error());
}
$result = mysql_query("select * FROM tablename");
if (!$result) {
echo 'Could not run query: ' . mysql_error();
exit;
}
$array= array();
while($row = mysql_fetch_array($result)) {
array_push($array, $row);
}
echo json_encode($array);
}

Minimal example:
index.html
$.getJSON( 'myphppage.php', {}, function(data){
// Do stuff here
});
myphpwebpage.php
echo json_encode($arrayname);

JSON Encode Failure

PHP PAGE:
<?php
include "linkpassword.inc";
function showVotes()
{
$showresult = mysql_query("SELECT * from mms") or die("Invalid query: " . mysql_error());
$row = mysql_fetch_assoc($showresult);
}
function addVote()
{
$sql= "UPDATE mms SET votes = votes+1 WHERE color = '".$_POST['color']."'";
$result= mysql_query($sql) or die(mysql_error());
return $result;
}
addVote();
showVotes();
?>
I am trying to get the output of the array to load into a JavaScript page where I can break up the array into seperate divs that have IDs assigned to them. Here is what I tried
<script>
$(document).ready(function () {
$('.answer').click(function (e) {
var color = $(this).attr("data-color");
$.ajax({
type: 'POST',
url: 'mm.php',
data: { color: color},
dataType: 'json',
cache: false,
success: function(showVotes) {
$('#rvotes').html(showVotes[0]);
},
error: function (jqXHR) {
}
})
})
});
</script>
Where am I going wrong??

From what you've posted in comments, what you have is an array of objects.. not html, as your function seems to indicate. Depending on what you want done, the answer would be either of the following, to access that object's properties:
showVotes[0].votes
Or
showVotes[0]['votes']
Eg:
$('#rvotes').html(showVotes[0].votes);
Or etc.
Second attempt:
Firstly, change your current 'showVotes' function to this:
function showVotes()
{
$showresult = mysql_query("SELECT * from mms") or die("Invalid query: " . mysql_error());
while ($row = mysql_fetch_assoc($showresult)) {
$response[] = $row;
}
return json_encode($response);
}
Secondly, remove your 'connected successfully' text from the page, as well as any other text generated by anything else(aka, the other function which returns a result pointer). I may be wrong, but it would seem to me that the generation of this other text is causing the returned json to be interpreted as malformed.
Quick explanation on PDO:
try {
$dbh = new PDO("mysql:host=localhost;dbname=dbname", "user", "password");
$dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
} catch (\PDOException $e) {
echo "Error! Could not connect to database: " . $e->getMessage() . "<br/>";
die();
}
Connecting to the database.. This is how I've learned to do it, though I've been warned(and downvoted) to not check for errors this way, though it was never explained why.
Database interaction:
$stmt = $dbh->prepare("UPDATE mms SET votes = votes+1 WHERE color = :color");
$stmt->bindParam(":color",$_POST['color']);
$stmt->execute();
Result use:
while ($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
$response[] = $row;
}
And so on and so forth. PDO escapes the values for you, so you don't have to worry about injection attacks.