MySQLデータベースから行をフェッチできません

Question

mysqlに少し錆びています。特にmysqlの場合はi ...下のスクリプトでは何も起こらず、変数$ row [username]などは空です。私は間違って何をしていますか？

<?php 

session_start(); 
include_once('./db_config.php'); 

if (mysqli_connect_errno()) 
    { 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
    } 

$username = mysqli_real_escape_string($conn, $_POST['username']); 
$password = md5(mysqli_real_escape_string($conn, $_POST['password'])); 

$query = "SELECT * FROM players WHERE username = '".$username."' AND password = '".$password."'"; 

if($res = mysqli_query($conn, $query)) 
{ 
    $row = mysqli_num_rows($res); 
    $_SESSION['Username'] = $row['username']; 
    $_SESSION['uID'] = $row['id']; 
    $_SESSION['Join_Date'] = $row['join_date']; 
    mysqli_free_result($res); 

    header('Location: ../index.php'); 
} else 
{ 
    echo mysqli_error($conn); 
} 

?> 

Answer 1

あなたはmyql_num_rowsを行っている代わりにレコードをフェッチしていない、これは、レコード

$row = mysqli_fetch_array($res,MYSQLI_ASSOC);

<?php 

session_start(); 
include_once('./db_config.php'); 

if (mysqli_connect_errno()) 
    { 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
    } 

$username = mysqli_real_escape_string($conn, $_POST['username']); 
$password = md5(mysqli_real_escape_string($conn, $_POST['password'])); 

$query = "SELECT * FROM players WHERE username = '".$username."' AND password = '".$password."'"; 

if($res = mysqli_query($conn, $query)) 
{ 

    $row = mysqli_fetch_array($res,MYSQLI_NUM); 
    $_SESSION['Username'] = $row['username']; 
    $_SESSION['uID'] = $row['id']; 
    $_SESSION['Join_Date'] = $row['join_date']; 
    mysqli_free_result($res); 

    header('Location: ../index.php'); 
} else 
{ 
    echo mysqli_error($conn); 
} 

?> 

にフェッチし

使用（これは、レコードの唯一の一切を与えません）