首先，如果我没记错，则将
sqlsrv_connect结果存储 到其中，
$conn并且该结果不是类obj它的资源，因此请删除
$db->conn
此示例将连接，然后从中获取返回的资源
sqlsrv_query
$conn_array = array ( "UID" => "sa", "PWD" => "root", "Database" => "nih_bw",);$conn = sqlsrv_connect('BILAL', $conn_array);if ($conn){ echo "connected"; if(($result = sqlsrv_query($conn,"SELECt * FROM routines")) !== false){ while( $obj = sqlsrv_fetch_object( $result )) { echo $obj->colName.'<br />'; } }}else{ die(print_r(sqlsrv_errors(), true));}
