php - 尝试发送UPDATE查询时，无法将mysqli_result类的对象转换为字符串

Question

我正在尝试只向表发送UPDATE查询，以为用户设置一个值，


  可捕获的致命错误：在第19行的E：\ web \ nginx \ html \ b1is \ profile.php中，类mysqli_result的对象无法转换为字符串


我不希望返回任何值，除非可以进行查询。

码：

<?php
ini_set('display_errors', 1); ini_set('display_startup_errors', 1); error_reporting(E_ALL);
session_start();
$sqlcon = include_once 'dbconnect.php';

if(!isset($_SESSION['user']))
{
 header("Location: login.php");
}
$res=mysqli_query($sqlcon,"SELECT * FROM users WHERE user_id=".$_SESSION['user']);
$userRow=mysqli_fetch_array($res);

if(isset($_POST['updprof']))
{
    $steamid = mysqli_real_escape_string($sqlcon,$_POST['steamid']);
    $discordid = mysqli_real_escape_string($sqlcon,$_POST['discordid']);
    if(isset($steamid))
    {
        if(!mysqli_query($sqlcon,"UPDATE users SET steamid = '$steamid' WHERE user_id = $res"))
        {
            ?> <script>alert("Update couldn't be made")</script> <?php
        }
    }

    if(isset($discordid))
    {
        if(!mysqli_query($sqlcon,"UPDATE users SET discordid = '$discordid' WHERE user_id = $res"))
        {
            ?> <script>alert("Update couldn't be made")</script> <?php
        }
    }
}
?>

Answer 1

$res=mysqli_query($sqlcon,"SELECT * FROM users WHERE user_id=".$_SESSION['user']);
^^^---mysqli object

    if(!mysqli_query($sqlcon,"UPDATE users SET steamid = '$steamid' WHERE user_id = $res"))
                                                                  string context----^^^

您不能将mysql结果对象/句柄直接传递给另一个查询。您必须先从结果中获取一个值，然后再使用该值。