I'm trying to sort out the names from the database based on the first letter by using (addr WHERE LIKE 'A%'). This gives me the following error message:

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in...

Does anyone know what’s wrong here? Everything works great when I do not use (addr WHERE LIKE'A%') in the query string.

require_once ('../mysqli_connect.php'); 

$q = "SELECT CONCAT(adresse) AS addr FROM gate_vei WHERE addr LIKE 'A%'"; 
$r = @mysqli_query ($dbc, $q); 

echo '<table align="center" cellspacing="3" cellpadding="3" width="75%">
	<tr><td align="left"><b>Adresse</b></td></tr>';

	while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)) {
		echo '<tr><td align="left">' . $row['addr'] . '</td>

	echo '</table>'; 

mysqli_free_result ($r); 	

addr is not a column name. You should probably use addresse

Yes, it solved the problem. Thank you pritaeas!