Failed to assign variable taken from database to json array - php

Question

I try to get the data from database to display data via ajax but failed to worked. It's partially working because data from mysql make this thing failed to function.
Here is my funds_transfer_backend.php page. This page will assign variable to json array.
session_start();
if(!isset($_SESSION['myusername']))
{
header("Location: ../index.html");
die();
}
include("../connect.php");
$myusername = $_SESSION['myusername'];
$sql="SELECT client_id FROM `client` WHERE username='$myusername'";
$result=mysqli_query($conn, $sql);
while ($row=mysqli_fetch_row($result)){
$id = $row['0'];
}
$index_num = $_POST['index_num'];
$to_account_num = $_POST['recipient'];
$amount = $_POST['amount'];
if ($amount == '' || $to_account_num == '' || $index_num == -1){
//echo "Please complete the form!";
$response = -1;
}
else {
// check account number exist
$query2 = "SELECT 1 FROM account WHERE id='$to_account_num' LIMIT 1";
if (mysqli_num_rows(mysqli_query($conn, $query2))!=1) {
//echo "Recipient account number is invalid!";
$response = -2;
}
else {
$query2 = "SELECT client.name, client.email FROM account JOIN client USING (client_id) WHERE account.id = '$to_account_num' LIMIT 1";
$result=mysqli_query($conn, $query2);
while($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
$name = $row['name'];
$email = $row['email'];
}
$response = 1;
}
} // check account num else bracket
$display = array('response' => $response, 'name' => $name);
echo json_encode($display);
However if I remove 'name' => $name from array the #stage div will trigger like image below:
Here is my funds_transfer.php page
<script type="text/javascript">
function update() {
var two = $('#index_num').val();
var three = $('#recipient_box').val();
var five = $('#amount_box').val();
$.post("funds_transfer_backend.php", {
index_num : two,
recipient : three,
amount : five
},function(data){
if (data.response==-1) {
$('#stage').show().html("Please complete the form!");
}
$('#stage').delay(2000).fadeOut();
},"json");
}
</script>
...other code goes here
<div id="stage" style="background-color:#FF6666; padding-left:20px; color: white;"></div>
<div id="confirm" style="background-color:#FF7800; padding-left:20px; color: white;"></div>
I try to check the data from db whether it exist using manual form method="post" and I can see the name being echo. Any help is appreciated and thanks in advance.

Answer 1

When your response is -1, your $name variable is undefined. So php could show a warning (depending on your settings) and you are trying to add an undefined variable to your array. This will invalidate your output / json.
You can set for example:
$name = '';
at the start of your script or check whether the variable is set with isset($name) before you try to use it to avoid these problems.
There are of course other solutions, like outputting your -1 directly and exiting the script there.

Answer 2

I always initialize my variables.
$myusername = isset($_SESSION['myusername']) ? $_SESSION['myusername'] : false;
Then you can safely do:
if ($myusername) {} without throwing warnings.
I do this weather I get my data from a db, post/get/session or json/ajax.
It takes a little extra time upfront but removes dozens of errors in the back end so you net more time.