Question

我目前正在尝试使用php和Mysql数据库创建一个网站。我尝试使用以下标记语言和PHP代码将我的网站连接到mysql数据库。但是它一直显示以下错误消息：

Notice: Undefined index: fname in C:\xampp\htdocs\signup.php on line 11
Notice: Undefined index: lname in C:\xampp\htdocs\signup.php on line 11
Notice: Undefined index: email in C:\xampp\htdocs\signup.php on line 11
Notice: Undefined index: password in C:\xampp\htdocs\signup.php on line 11
Notice: Undefined index: cpassword in C:\xampp\htdocs\signup.php on line 11

这是我的代码：

    <?php
    $con=mysqli_connect("localhost","root","p11","daot");
    // Check connection
    if (mysqli_connect_errno())
      {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
      }

    $sql="INSERT INTO registration (FirstName, LastName, EmailAddress,Password,ConfirmPassword)
    VALUES
    ('$_POST[fname]','$_POST[lname]','$_POST[email]','$_POST[password]','$_POST[cpassword]')";

    if (!mysqli_query($con,$sql))
      {
      die('Error: ' . mysqli_error($con));
      }
    echo "1 record added";

    mysqli_close($con);
    ?> 

<html>
<head>
<link rel="stylesheet" type="text/css" href="mastercss.css">
<title>SIGN UP PAGE</title>
</head>
<body>
<?php include 'header.php'; ?>
<div class="leftbar">
 </div>
<div class="content">
<h1 class="h1">complete the following form to register</h1>
<fieldset style="width:450px; background:gray;">
<form autocomplete="on" method="post" action="insert.php">
  <label for="fname">First Name:</label>
  <input type="text" name="fname" required><br><br>
  <label for="lname">Last Name:</label>
  <input type="text" name="lname"><br><br>
  <label for="email">Email:</label>
  <input type="email" name="email"><br><br>
  <label for="password">Password:</label>
  <input type="password" name="password"><br><br>
  <label for="cpassword">Confirm Password</label>
  <input type="password" name="cpassword"><br><br>
  <!--<label for="sex">Sex</label><input type="radio" name="sex" value="female"> Female
  <input type="radio" name="sex" value="male">Male<br>
  <label for="select">Birthday</label>
<select name="birthday_Month" id="month">
<option value="0" selected="1">Month</option>
<option value="1">January</option>
<option value="2">February</option>
<option value="3">March</option>
</select>
<select name="birthday_day" id="month">
<option value="0" selected="1">Day</option>
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
</select>
<select name="birthday_year" id="year">
<option value="0" selected="1">year</option>
<option value="2010">2010</option>
<option value="2011">2011</option>
<option value="2012">2012</option>
</select><br><br>-->
<input type="submit" value="SIGN UP" style="width:100: height:100" name="Submit">
</form>
</fieldset>
</div>
<div class="rightbar"><br><br>
<a href="https://www.twitter.com"><img src="tw1.jpg">
<a href="https://www.facebook.com"><img src="fb2.jpg"></a>
</div>
<?php include "footer.php";?>

</body>
</html>

我的代码出了什么问题？

Answer 1

因为，如果你只按提交按钮，它会发布值，除此之外没有帖子值。这就是你得到未定义索引错误的原因，因此添加isset($_POST['Submit'])将解决问题

   if(isset($_POST['Submit'])){     

    $sql="INSERT INTO registration (FirstName, LastName, EmailAddress,Password,ConfirmPassword)
    VALUES
    ('".$_POST['fname']."','$_POST[lname]','$_POST[email]','$_POST[password]','$_POST[cpassword]')";

    if (!mysqli_query($con,$sql))
    {
        die('Error: ' . mysqli_error($con));
    }
    echo "1 record added";

    mysqli_close($con);

   }

Answer 2

第一次加载此页面时，它没有发布数据（$ _ POST [fname]，$ _ POST [lname] ...）。所以它给你这个通知。

您应该检查您的服务器请求方法如果它是帖子，那么你有发布数据，你可以运行你的PHP代码。

if ($_SERVER['REQUEST_METHOD'] === 'POST') {
$con=mysqli_connect("localhost","root","p11","daot");
    // Check connection
    if (mysqli_connect_errno())
      {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
      }

    $sql="INSERT INTO registration (FirstName, LastName, EmailAddress,Password,ConfirmPassword)
    VALUES
    ('$_POST[fname]','$_POST[lname]','$_POST[email]','$_POST[password]','$_POST[cpassword]')";

    if (!mysqli_query($con,$sql))
      {
      die('Error: ' . mysqli_error($con));
      }
    echo "1 record added";

    mysqli_close($con);

}

php Mysql未定义索引错误

2 个答案: