用PHP准备的MySQLi语句

抱歉,如果我对此感到陌生.

        if(!isset($stmt_counttime)) { $stmt_counttime = $MysqLi->prepare("SELECT count(1) as count FROM `stop_times` where `stop_id` = ? and `trip_id` = ? LIMIT 1");
    $stmt_counttime->bind_param('ss',$stop_id,$trip_id); }
    $stop_id = $data[3];
    $trip_id = $data[0];
    $stmt_counttime->execute();
    $stmt_counttime->bind_result($count_time_result);
    $stmt_counttime->fetch();

    if(!isset($stmt_counttrip)) { $stmt_counttrip = $MysqLi->prepare("SELECT count(1) as count FROM `trips` where `trip_id` = ?");
    $stmt_counttrip->bind_param('s',$tripid);
     }

    $tripid = $data[0];
    $stmt_counttrip->execute();
    $stmt_counttrip->bind_result($count_trip_result);
    $stmt_counttrip->fetch();

第67行：$stmt_counttrip-> bind_param(‘s’,$tripid);

错误：致命错误：在第67行的/home/fallouta/public_html/ocbus/google_transit/stop_times.PHP中的非对象上调用成员函数bind_param()

如果有人能让我知道,我在做什么错？请告诉我.

注意：我需要trip_id的值作为字符串而不是整数.

谢谢,
赛勒斯

解决方法:

我意识到我的错误.其中应包括以下声明：

$stmt-> store_result();

因为您不能同时有两个查询.随时添加更多信息.