The issue is that you’re not saving the mysqli connection. Change your connect to:
$aVar = mysqli_connect(‘localhost’,’tdoylex1_dork’,’dorkk’,’tdoylex1_dork’);
And then include it in your query:
$query1 = mysqli_query($aVar, “SELECT name1 FROM users
ORDER BY RAND()
LIMIT 1″);
$aName1 = mysqli_fetch_assoc($query1);
$name1 = $aName1[‘name1’];
Also don’t forget to enclose your connections variables as strings as I have above. This is what’s causing the error but you’re using the function wrong, mysqli_query returns a query object but to get the data out of this you need to use something like mysqli_fetch_assoc http://php.net/manual/en/mysqli-result.fetch-assoc.php to actually get the data out into a variable as I have above.
the mysqli_queryexcepts 2 parameters , first variable is mysqli_connectequivalent variable , second one is the query you have provided
$name1 = mysqli_connect(localhost,tdoylex1_dork,dorkk,tdoylex1_dork);
$name2 = mysqli_query($name1,”SELECT name FROM users ORDER BY RAND() LIMIT 1”);