$sql = 'SELECT ALL l.dpt, COUNT(*) as total FROM localisation AS l LEFT OUTER JOIN article AS a ON l.dpt = a.departement GROUP BY 1 ORDER BY 1 ASC';...
$sql = 'SELECT ALL l.dpt, COUNT(*) as total FROM localisation AS l LEFT OUTER JOIN article AS a ON 1.dpt = a.departement GROUP BY 1 ORDER BY 1 ASC';...