mysqli inserts typed values but not variables - variables

Question

I have tried to get the following code to work. It is an ajax-call where I send a json-encoded string 'data':
<?php
require_once('connect.php'); // get values for mysqli-connect
$arr1 = json_decode($_GET['data']);
$arr = array_values($arr1);
if ($GLOBALS['TSFE']->loginUser) {
$mysqli = new mysqli($server,$user,$pw,$db);
if ($mysqli->connect_error) {
die('Error : ('. $mysqli->connect_errno .') '. $mysqli->connect_error);
}
//values to be inserted in database table
$var1 = "Hello World";
$var2 = 45;
$var3 = "an encoded array";
/*
$var1 = $arr[1];
$var2 = $arr[2];
$var3 = json_encode($arr[3]);
*/
$insert_row = $mysqli->query("INSERT INTO table_testing
(event, pid, myarray) VALUES ('$var1', '$var2', '$var3')");
if($insert_row){
print 'Success! ID of last inserted record is : ' .$mysqli->insert_id .'<br />';
}else{
die('Error : ('. $mysqli->errno .') '. $mysqli->error);
}
} else {
echo "User is NOT logged in!\n";
}
?>
My problem is that this code works! It works fine with the three values that are stored in $var1-3. They are saved in the correct table.
But - if I try to store the out-commented three values from $arr[1-3] into $var[1-3] it doesn't work? The three values saved are ' ' (empty), '0' and 'NULL'.
I have tried to put these lines in right after $arr is declared:
echo $arr[1]."\n";
echo $arr[2]."\n";
echo json_encode($arr[3])."\n";
In the console the correct values are printed, so they are valid and exist. It is a string, an integer and an array.
What is it I have overlooked or not understood?

Answer 1

Indexes of PHP arrays start are 0-based.
Try with this code:
echo $arr[0] . "\n";
echo $arr[1] . "\n";
echo json_encode($arr[2]) . "\n";