Php 警告：mysql_num_rows（）：提供的参数不是有效的mysql结果资源-如何修复它？

我有一个脚本，应该添加蒸汽按钮登录。 它实际上显示了按钮，但出现了一个错误，并且登录不起作用。这是我的代码：

<?php
    /*
    DATABASE INFO FOR steamids
    1   userid  int(11) //when u have login and register system, here u put user's id
    2   steamid varchar(30)
    3   profileurl  text
    4   nickname    varchar(50)
    5   name    varchar(50) 
    6   avatar  text
    */
    $api = "xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx";
    include 'scripts/libraries/openid.php';

    mysql_connect("localhost","user","password");
    mysql_select_db("database"); //database name

    $my_id = "kartm_user"; //user's id
    $OpenID = new LightOpenID("localhost");

    if(isset($_GET['disconnect'])) {
        mysql_query("DELETE FROM steamids WHERE userid='$my_id'");
    }

    $query = mysql_query("SELECT * FROM steamids WHERE userid='$my_id'");
    $rows = mysql_num_rows($query);
    if($rows == 0) {
        if(!$OpenID->mode) {
            $OpenID->identity = "https://steamcommunity.com/openid/";
            $logmein = $OpenID->authUrl();

            echo "<a href=$logmein><img src='http://steamcommunity-a.akamaihd.net/public/images/signinthroughsteam/sits_large_noborder.png' /></a>";

        } elseif($openid->mode == 'cancel') {
            echo 'User has canceled authentication!';
        } else {
            $_SESSION['T2SteamAuth'] = $OpenID->validate() ? $OpenID->identity : null;
            $_SESSION['T2SteamID64'] = str_replace("http://steamcommunity.com/openid/id/", "", $_SESSION['T2SteamAuth']);
            if($_SESSION['T2SteamAuth'] !== null) {
                $steamidonly = $_SESSION['T2SteamID64'];
                $query = mysql_query("INSERT INTO steamids VALUES('$my_id','$steamidonly','','','','')");
            }
            header("Location: login.php"); //change it to your file
        }
    } else {
        $resultid = mysql_result($query, 0, 'steamid');

        $getcontent = file_get_contents("http://api.steampowered.com/ISteamUser/GetPlayerSummaries/v0002/?key=$api&steamids=$resultid");
        $data = json_decode($getcontent);
        $user_steamid = $data->response->players[0]->steamid;
        $personaname = $data->response->players[0]->personaname;
        $profileurl = $data->response->players[0]->profileurl;
        $avatarfull = $data->response->players[0]->avatarfull;
        $realname = $data->response->players[0]->realname;
        mysql_query("UPDATE steamids SET profileurl='$profileurl', nickname='$personaname', name='$realname', avatar='$avatarfull'");

        //EXAMPLE
        echo "<a href='$profileurl'><img src='$avatarfull' width='50px'></a>
        $personaname<br>
        <a href='?disconnect'><input type='button' value='DISCONNECT'></a> ";
    }

?>

这是我的错误：

警告：mysql_num_rows:在第25行的/home/kartm/public_html/scripts/streamlogin.php中，提供的参数不是有效的mysql结果资源

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如何修复它？

这里@哈弗：是的，我看到了，但我不知道如何修复我的代码。请将相关代码添加到你的问题正文中。例如，我是移动的&不希望有不必要的点击来查看代码调用mysql\u查询失败-可能查询无效，或者您的连接未打开。尝试回显SQL的含义，并在数据库客户机中运行它。回显mysql\u错误；每次mysql之后_