Question

我正在尝试检查登录，然后尝试回显基本用户信息。在此先感谢大家的帮助。有关为什么我无法根据特定登录用户回显行数据的任何建议？

     <?php
        <?php
    session_start();
    // dBase file
    include "config.php";
ini_set('display_errors', 1);
    if (isset($_GET["op"] ) && $_GET["op"] == "login")  
  {
  if (!$_GET["username"] || !$_GET["password"])
    {
    die("You need to provide a username and password.");
    }

  // Create query
  $q = "SELECT * FROM `users` "
    ."WHERE `username`='".$_GET["username"]."' "
    ."AND `password`=PASSWORD('".$_GET["password"]."') "
    ."LIMIT 1";
  // Run query
  $r = mysql_query($q);
  $row = mysql_assoc_array($r);
  $count = mysql_num_rows($r);
  if ($obj = mysql_fetch_object($r) && $count == 1)
    {
    // Login good, create session variables
    $_SESSION["id"] = $row['id'] = $obj->id;
    $_SESSION["user"] = $row['username'] = $obj->username;
    $_SESSION["time"] = time();
    // Redirect to member page
    Header("Location: welcome.php");
    }

  else
    {
    // Login not successful
    die("Sorry, could not log you in. Wrong login information.");
    header("Location: login2.php");
    }
}
     {
//If all went right the Web form appears and users can log in
  echo "<form action=\"?op=login\" method=\"GET\">";
  echo "Username: <input name=\"username\" name=\"username\" size=\"32\"><br />";
  echo "Password: <input type=\"password\" name=\"password\" size=\"32\"><br />";
  echo "<input type=\"submit\" value=\"Login\">";
  echo "</form>";
  }
    ?>

Answer 1

您正在使用mysql_fetch_object，因此您需要执行以下操作：

$_SESSION["user"] = $obj->username;

而不是：

$_SESSION["user"] = $row['username'];

登录脚本不起作用

1 个答案: