Question

这些是我的代码我继续从php端获得未定义的索引，但我想我已经清楚地从表单方面发布了正确的信息，但仍然...... 帮我PLS !!!帮我PLS !!!帮我PLS !!!帮我PLS !!!帮我PLS !!!

<form class="modal-content animate" action="TenantSdb.php">
<div class="container">
  <label><b>Username</b></label>
  <input type="text" placeholder="Enter Username" name="uname" required>

  <label><b>Password</b></label>
  <input type="password" placeholder="Enter Password" name="psw" required>

  <label><b>Email address</b></label>
  <input type="text" placeholder="Enter Email" name="email" required>

  <label><b>Contact No</b></label>
  <input type="text" placeholder="Enter Contact" name="contact_no" required>

  <label><b>Looking for which area</b></label>
  <input type="text" placeholder="Enter Area" name="area" required>

  <label><b>Gender</b></label>
  <input type="text" placeholder="Enter Gender" name="gender" required>

  <label><b>Age</b></label>
  <input type="text" placeholder="Enter Age" name="age" required>

  <label><b>Max Budget</b></label>
  <input type="text" placeholder="Enter Budget" name="max_budget" required>

  <label><b>Staying With</b></label>
  <input type="text" placeholder="Enter Staying with" name="staying_with"
  required>

  <label><b>Race</b></label>
  <input type="text" placeholder="Enter Race" name="race" required>



  <button type="submit">Sign Up</button>
  <input type="checkbox" checked="checked"> Remember me
</form>
</div>

`

$servername ="localhost";
$dbusername ="root";
$dbpassword ="";
$dbname = "tenantsdb";

$uname          =   $_POST['uname'];
$psw            =   $_POST['psw'];
$email          =   $_POST['email'];
$contact_no     =   $_POST['contact_no'];
$area           =   $_POST['area'];
$gender         =   $_POST['gender'];
$age            =   $_POST['age'];
$max_budget     =   $_POST['max_budget'];
$staying_with   =   $_POST['staying_with'];
$race           =   $_POST['race'];


$conn = new mysqli($servername, $dbusername,$dbpassword, $dbname);

if ($conn->connect_error)
{
die("Connection failed : " . $conn->connect_error);
}

$sql   = "INSERT INTO tenantsignup 
(uname,psw,email,contact_no,area,gender,max_budget,staying_with,race) VALUES 

('$uname','$psw','$email','$contact_no','$area','$gender','$max_budget',
'$staying_with','$race')";

if($conn->query($sql) === TRUE)
{
echo "Thank you !";
}

else
{
echo "error" . $sql . "<br>" . $conn->error;
}



$conn->close();



?> 

i am getting undefined index from 
$uname          =   $_POST['uname'];
$psw            =   $_POST['psw'];
$email          =   $_POST['email'];
$contact_no     =   $_POST['contact_no'];
$area           =   $_POST['area'];
$gender         =   $_POST['gender'];
$age            =   $_POST['age'];
$max_budget     =   $_POST['max_budget'];
$staying_with   =   $_POST['staying_with'];
$race           =   $_POST['race'];


Keep gettting undefined index ^^^ from top, can anyone tell me the problem ?
Please help me, thank you very much

Answer 1

您已指定属性method='POST'

<form class="modal-content animate" action="TenantSdb.php" method="POST">

默认方法是GET。因此，如果您未指定方法，则只能使用$_GET

来访问它

Answer 2

您必须指定方法，否则默认为 GET 方法。如果您不知道表单中使用的确切方法，则使用$ _REQUEST。 $ _REQUEST可以发布数据以及 GET 数据。它适用于两者。

将信息从表单发布到xampp时未定义的索引

2 个答案: