Notice: Trying to get property 'num_rows' of non-object in [duplicate] - php

Question

This question already has an answer here:
What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such
(1 answer)
Closed 2 years ago.
<?php
$queryuser = "SELECT id, email, firstname, lastname, address, city, country FROM users WHERE role = 'client'";
$resultuser = $connection->query($queryuser);
if ($resultuser->num_rows > 0)
{
// output data of each row
while ($rowuser = $resultuser->fetch_assoc())
{
$id_user = $rowuser['id'];
$firstname = $rowuser['firstname'];
$lasttname = $rowuser['lastname'];
$email = $rowuser['email'];
$phone = $rowuser['phone'];
$city = $rowuser['city'];
$state = $rowuser['state'];
$address = $rowuser['address'];
?>

Answer 1

Perhaps you've copied code from this page https://www.w3schools.com/php/php_mysql_select.asp
Please understand code line by line first. You have not define $connection variable in your code (may be it is a intentional approach for security).
Your notice saying that you are trying to get num_rows property from the object $resultuser but in this query $resultuser is not a object due to any error/faults of your query because query() returns FALSE on failure and object on success.
To confirm, you can check your result set by using var_dump();
Please replace this line if $resultuser->num_rows > 0 with (!empty($resultuser->num_rows) && ($resultuser->num_rows > 0)))