compare value in mysqli query

Question

I have one table called tbl_favorite. there two field called user_id and quote_id. I am passing this both field in parameter. I want check that user_id have any row with that quote_id. if its exist I want response like "already exist" else I want insert that data in table.

    <?php
header('Content-Type: application/json; Charset=UTF-8');
include("dbconnection.php");
$userid= $_GET['userid'];
$quoteid= $_GET['quoteid'];

// I am confused to write condition
$query= mysqli_query($conn,"SELECT fav_id FROM tbl_fav  WHERE  user_id='".$userid.'" AND fav_qu_id="'.$quoteid.'"");

if(mysqli_num_rows($query) > 0){

$response="already exist";

}
else{
$query= mysqli_query($conn,"INSERT INTO tbl_fav(user_id,fav_qu_id VALUE('".$userid."','".$quoteid."')");
$response['message']='success';

}
echo json_encode($response);
?>

my url is like below

example.com/api.php?user_id=1&&quote_id=1

I am getting error called Parse error: syntax error, unexpected '"'. Let me know if someone correct me. I am android developer and does not know proper PHP. Thanks

Answer 1

$resource = "SELECT * FROM tbl_favorite WHERE user_id='{$userid}' AND quote_id='{$quoteid}'";
$result = if (mysqli_query($query)){

    return true;

}

if ($result){
  echo "that quote exists within the given user";
}