Question

所以我第一次尝试使用COOKIES失败了。我正在尝试完成REMEMBER ME功能。我在youtube上观看了几个视频，并尝试了Stack Overflow上的几种方法。一定是我......

在我的INDEX.PHP上，我有这个登录表单：

 <form action="index.php" method="get" id="loginForm" name="loginForm">
 <input type="text" id="username" name="username" placeholder="Username" />
 <input type="password" id="password" name="password" placeholder="Password"/>
 <input type="checkbox" id="rememberMe" name="rememberMe" /> Remember Me
 <button type="button" id="loginSubmit" name="loginSubmit">Sign In</button>
 </form>

在同一页上，靠近底部，我有一些JQUERY：

 $('#loginSubmit').on('click', function()
 {
   var username = $('#username').val();
   var password = $('#password').val();
   var rememberMe = $('#rememberMe').prop('checked');

   // some validation to make sure no field is blank
   // then post to a PHP script

   $.post('api/checkLogin.php', {username: username, password: password, rememberMe: rememberMe}, function(data)
   {
     if(data == 0)
     {
       alert('username and password do not match');
       return false;
     }
     else
     {
       window.location.href = "main.php";
     }
   });

checkLogin.php脚本如下所示（尽可能缩短）：

 <?php
   include("../include/database.php");
   include("../include/sessions.php");

   if($_POST['username'] == true || $_POST['password'] == true)
   {
     $username = $_POST['username'];
     $password = $_POST['password'];
     $rememberMe = $_POST['rememberMe'];

     $select = "SELECT username, password FROM table WHERE username = '$username'";
     $query = mysqli_query($dbc, $select);
     $row = mysqli_fetch_array($query);

     $dbusername = $row['username'];
     $dbpassword = $row['password'];

    if($dbpassword == $password)
    {
      $_SESSION['username'] = $username;
      // here's where I'm trying to set the cookie
      if($rememberMe == "True")
      {
        $cookie_username = $username;
        setcookie('username', $cookie_username, time()+(86400*7));
      }
      // send json if username/password matched
      $out = 1;
      echo json_encode($out);
      mysqli_free_result($query);       
    }
    else
    {
      $out = 0;
      echo json_encode($out);
      mysqli_free_result($query);
    }
   }
 ?>

当我成功地被重定向到main.php时，我试图回应COOKIE无济于事：

 <?php
   echo "cookie name is " . $_COOKIE['username'];
 ?>

当我使用Chrome的检查器时，它只显示：

 "cookie name is        "

我做错了什么？

PHP没有COOKIES的乐趣

0 个答案: