PHP Inserting variables in array, but can't json_encode - php

Question

Im with this for days, i tested if variables are being filled and they are, but i think somehow im not doing array right because 'echo json_encode($response)' is printing nothing. I also tried to replace variables in array to just random numbers and everything went fine, just with variables does work.
public function getPlaces() {
$places = array();
$stmt = "SELECT * FROM poi ";
$retval = mysqli_query( $this->conn, $stmt );
if ($retval) {
while($row=mysqli_fetch_array($retval)) {
$name = $row["name"];
$lat = $row["latitude"];
$lng = row["longitude"];
$desc = $row["descricao"];
$rating = $row["rating"];
$lm = $row["lm"];
$la = $row["la"];
$lv = $row["lv"];
$places[] = array('name'=> $name, 'lat'=> $lat, 'lng'=> $lng, 'desc'=> $desc, 'rating'=> $rating,
'lm'=> $lm, 'la'=> $la, 'lv'=> $lv);
}
return $places;
}else
return false;
}
The function is called here:
require_once 'include/DB_Functions.php';
$db = new DB_Functions();
$response = array("error" => FALSE);
$places = $db->getPlaces();
if ($places != false) {
$response["places"] = $places;
echo json_encode($response);
} else {
$response["error"] = TRUE;
$response["error_msg"] = "Error";
echo json_encode($response);
}

Answer 1

Instead of using associative array, use mysqli_fetch_object() to get the result set as object. You do not have to type all the field names if you use array_push().
while($row=mysqli_fetch_object($retval)):
array_push($places, $row);
endwhile;
Vardump the $places array to check the array.
E.g.
vardump($places); die;
or
print_r($places); die;

Answer 2

Did you notice the $ is missing before the $row variable on that line ?
$lng = row["longitude"];
This kind of missed (bad handled) Notice + Warning won't make your array.
Notice also that if your array is filled with some type of value it won't json_encode. Especially NaN values (thinking of this because of the nature of some fields like "rating" that could result in a NaN value).
Try by steps :
Dump your result for one line only with a LIMIT 1 in your SQL query to see if it works. Try only with one field...
Notice that $places = [];
is more optimized than a
$places = array();
Same with
$places[] = ['name'=> $name, ... ];