Question

This code works, and outputs an array as intended, but throws the following error message:

PHP Notice: Trying to get property of non-object…

Here is my code:

$mysqli = new mysqli("localhost","user","pass","database");
$sql = "SELECT keyterm
    FROM keyterms";
$results = $mysqli->query($sql);
while($result = $results->fetch_object()->keyterm)
    $keyterms[] = $result;

Answer 1

fetch_object() will return null when there are no more results in your result set, so the last iteration of the while loop will attempt to access the keyterm property on a null value.

You should assign the result object to $result and then access the property in the loop body, e.g.:

while($result = $results->fetch_object())
    $keyterms[] = $result->keyterm;