PHP错误：注意：尝试获取非对象的属性

I am a PHP newbe. I'm trying to load an array from this function:

我是一名PHP新手。我正在尝试从此函数加载数组:

funcs.php (sql returns 1 record)

funcs.php(sql返回1条记录)

function loaduser($conn, $mysql_db, $x_email, $x_password) {
$sql = "SELECT userid, userf, userl
        FROM ".$mysql_db.".users 
        WHERE email = '".mysqli_real_escape_string($conn, $x_email)."' 
        AND pw = '".mysqli_real_escape_string($conn, $x_password)."'";
    //echo($sql);
    $Auser = $conn->query($sql);
    return $Auser;
}

loaduser.php

<?php require_once("../includes/funcs.php")

$x_email = $_GET['email'];
$x_password = $_GET['password'];
$Auser = loaduser($conn, $mysql_db, $x_email, $x_password);
var_dump($Auser);
if ($Auser[0]->num_rows > 0) {
    $row=mysqli_fetch_assoc($Auser);
    session_start();
    $_SESSION['userid'] = $row["userid"];
    echo "1";
} else {
    echo "0";
}
?>

but I get this error:

但我得到这个错误:

Notice: Trying to get property of non-object in loaduser.php on line 7

注意:尝试在第7行的loaduser.php中获取非对象的属性

Please help me understand the issue TQ

请帮我理解TQ的问题

EDIT - left $Auser[0] in by mistake should read $Auser->

编辑 - 错误地留下$ Auser [0]应该读$ Auser->

EDIT - this worked fine when in same file and not in a function

编辑 - 这在同一个文件中工作正常,而不是在函数中

EDIT - FIXED was db permissions issue - but no access denied error was shown

编辑 - 修复是数据库权限问题 - 但没有显示访问被拒绝错误

1 个解决方案

#1

$Auser as a query result is not an array, so this should work:

$ Auser作为查询结果不是一个数组,所以这应该工作:

if ($Auser->num_rows > 0) {
...
}