Notice: Trying to get property 'num_rows' of non-object in

Question

<?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'];
?>

Your query failed. Check for [mysqli errors](http://php.net/manual/en/mysqli.error.php) to find out why. — aynber, Jan 17 '19 at 16:38
There's a really big, common, canonical duplicate for this ... if only SO's search was a little better I'd find it in a couple of seconds... — CD001, Jan 17 '19 at 16:41
Welcome to SO. Please read [ask] and improve your question using the [edit](https://stackoverflow.com/posts/54240336/edit) link below the question. — Pinke Helga, Jan 17 '19 at 16:41

score 0 · Answer 1 · answered Jan 17 '19 at 19:48

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)))

Notice: Trying to get property 'num_rows' of non-object in

1 Answers1