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));
}
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