Question

在这里，我创建了一个包含登录和注册的表单。

当我运行我们的代码时，它会在消息中显示未定义的变量，并且无法在后端存储用户输入值。

以下是供您参考的完整代码：

db.php中：

<?php
$connection = mysqli_connect('localhost','root','','sign-log') or die(mysqli_error($connection));
?>

index.php（带css）：

<?php
include('db.php');
if(isset($_POST['action']))
{          
    if($_POST['action']=="login")
    {
        $email = mysqli_real_escape_string($connection,$_POST['email']);
        $password = mysqli_real_escape_string($connection,$_POST['password']);
        $strSQL = mysqli_query($connection,"select name from users where email='".$email."' and password='".md5($password)."'");
        $Results = mysqli_fetch_array($strSQL);
        if(count($Results)>=1)
        {
            $message = $Results['name']." Login Sucessfully!!";
        }
        else
        {
            $message = "Invalid email or password!!";
        }        
    }
    elseif($_POST['action']=="signup")
    {
        $name       = mysqli_real_escape_string($connection,$_POST['name']);
        $email      = mysqli_real_escape_string($connection,$_POST['email']);
        $password   = mysqli_real_escape_string($connection,$_POST['password']);
        $query = "SELECT email FROM users where email='".$email."'";
        $result = mysqli_query($connection,$query);
        $numResults = mysqli_num_rows($result);
        if (!filter_var($email, FILTER_VALIDATE_EMAIL)) // Validate email address
        {
            $message =  "Invalid email address please type a valid email!!";
        }
        elseif($numResults>=1)
        {
            $message = $email." Email already exist!!";
        }
        else
        {
            mysql_query("insert into users(name,email,password) values('".$name."','".$email."','".md5($password)."')");
            $message = "Signup Sucessfully!!";
        }
    }
}


echo '<script type="text/javascript" src="jquery-1.8.0.min.js"></script>
<link rel="stylesheet" href="http://code.jquery.com/ui/1.10.3/themes/smoothness/jquery-ui.css" />
<script src="http://code.jquery.com/ui/1.10.3/jquery-ui.js"></script>
<style type="text/css">
input[type=text]
{
  border: 1px solid #ccc;
  border-radius: 3px;
  box-shadow: inset 0 1px 2px rgba(0,0,0,0.1);
  width:200px;
  min-height: 28px;
  padding: 4px 20px 4px 8px;
  font-size: 12px;
  -moz-transition: all .2s linear;
  -webkit-transition: all .2s linear;
  transition: all .2s linear;
}
input[type=text]:focus
{
  width: 400px;
  border-color: #51a7e8;
  box-shadow: inset 0 1px 2px rgba(0,0,0,0.1),0 0 5px rgba(81,167,232,0.5);
  outline: none;
}
input[type=password]
{
  border: 1px solid #ccc;
  border-radius: 3px;
  box-shadow: inset 0 1px 2px rgba(0,0,0,0.1);
  width:200px;
  min-height: 28px;
  padding: 4px 20px 4px 8px;
  font-size: 12px;
  -moz-transition: all .2s linear;
  -webkit-transition: all .2s linear;
  transition: all .2s linear;
}
input[type=password]:focus
{
  width: 400px;
  border-color: #51a7e8;
  box-shadow: inset 0 1px 2px rgba(0,0,0,0.1),0 0 5px rgba(81,167,232,0.5);
  outline: none;
}
</style>  
  <script>
  $(function() {
    $( "#tabs" ).tabs();
  });
  </script>
</head>
<body>
 <b>'.$message.'</b>
<div id="tabs" style="width: 480px;">
  <ul>
    <li><a href="#tabs-1">Login</a></li>
    <li><a href="#tabs-2" class="active">Signup</a></li>

  </ul>                 
  <div id="tabs-1">
  <form action="" method="post">
    <p><input id="email" name="email" type="text" placeholder="Email"></p>
    <p><input id="password" name="password" type="password" placeholder="Password">
    <input name="action" type="hidden" value="login" /></p>
    <p><input type="submit" value="Login" /></p>
  </form>
  </div>
  <div id="tabs-2">
    <form action="" method="post">
    <p><input id="name" name="name" type="text" placeholder="Name"></p>
    <p><input id="email" name="email" type="text" placeholder="Email"></p>
    <p><input id="password" name="password" type="password" placeholder="Password">
    <input name="action" type="hidden" value="signup" /></p>
    <p><input type="submit" value="Signup" /></p>
  </form>
  </div>
</div>';
?>

db.sql：

CREATE TABLE `users` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(240) NOT NULL,
  `email` varchar(240) NOT NULL,
  `password` varchar(240) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=MyISAM AUTO_INCREMENT=3 DEFAULT CHARSET=latin1 AUTO_INCREMENT=3 ;

如何在后端存储输入值以及如何在消息中修复未定义的变量。

任何人都可以帮助我吗？

Answer 1

我似乎无法在评论中得到这一点，所以这里有更完整的解释：您正在将mysqli_*函数与已弃用的mysql_*函数混合使用。这与您收到的undefined variable警告无关。

问题在于：

    else
    {
        mysql_query("insert into users(name,email,password) values('".$name."','".$email."','".md5($password)."')");
    //  ^^^^^^^^^^^ not good
        $message = "Signup Sucessfully!!";
    }

mysqli_*和mysql_*函数不能像那样交换，你可以使用两者，但是你必须为每个函数打开一个数据库连接。

因此，要解决您的问题，您需要执行与上一个查询相同的操作：

$result = mysqli_query($connection,'insert ......');

Answer 2

变量$ message在IF体内，在此之前初始化它

$message = '';
if(isset($_POST['action']))
{.....

输入值没有存储在mysql数据库中

2 个答案: