提交时数据不保存按有（注意：未定义索引：第 18 行 C:\\xampp\\htdocs\\CW\\register.php 中的名字

Question

enter code here   
               <?php

//declare the basic variables
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "coursework";

//create connection
$link = new mysqli($servername, $username, $password, $dbname);

   if($link->connect_error){
      die ("connection failed: " . $link->connect_error);
   }


$sql = "INSERT INTO user (firstname, lastname, email, phn_number, password)                    VALUES ('".$_POST['firstname']."', '".$_POST['lastname']."',   '".$_POST['email']."', '".$_POST['phn_number']."', '".$_POST['password']."')";
 if (mysqli_query($link, $sql)){
   echo "Signed up Successfully";
 }else{
   echo "Error: Sign up Unsuccessfull $sql . " . mysqli_error($link);
 }
 //close db
 $link -> close();
  ?>

this is my PHP file code. Every time I run it creats a new user but the data entered by the user doesn't save in the database. and an error shows up

Notice: Undefined index: firstname in C:\xampp\htdocs\C.W\register.php on line 18
Notice: Undefined index: lastname in C:\xampp\htdocs\C.W\register.php on line 19
Notice: Undefined index: email in C:\xampp\htdocs\C.W\register.php on line 20
Notice: Undefined index: phn_number in C:\xampp\htdocs\C.W\register.php on line 21
Notice: Undefined index: password in C:\xampp\htdocs\C.W\register.php on line 22

help!!!!

Answer 1

update your code with this one:

 if($link->connect_error){
  die ("connection failed: " . $link->connect_error);
  }

$sql ="INSERT INTO user (firstname, lastname, email, phn_number, password) VALUES ('Dinanath', 'Thakur', 'dina@example.com', 123456789, 'xyz')";

your html form should be in the format, update your html form:

<p>
    <label for="firstname"> First Name: </label>
    <input type="text" name="firstname" id="firstname"> <br/> <br/>
    <label for="lastname"> Last Name: </label>
    <input type="text" name="lastname" id="lastname"> <br/> <br/>
    <label for="email"> E-mail: </label>
    <input type="text" name="email" id="email"> <br/> <br/>
    <label for="phn_number"> Phone Number: </label>
    <input type="text" name="phn_number" id="phn_number"> <br/> <br/>
    <label for="password"> Password: </label>
    <input type="text" name="password" id="password">
</p>
<input type="submit" value="Sign Up">

Answer 2

You might forgot to include this:

form action="name_of_your_php_file.php" method="POST"

Inserting this to the code you provided above gives us:

<form action="name_of_your_php_file.php" method="POST">
<p> <label for "firstname"> First Name: </label>
<input type="text" name"firstname" id="firstname"> <br/ > <br/ > 
<label for "lastname"> Last Name: </label> <input type="text" name"lastname" id="lastname"> <br/ > <br/ > 
<label for "email"> E-mail: </label> <input type="text" name"email" id="email"> <br/ > <br/ >
<label for "phn_number"> Phone Number: </label> <input type="text" name"phn_number" id="phn_number"> <br/ > <br/ > <label for "password"> Password: </label>
<input type="text" name"password" id="password"> </p>
<input type="submit" value="Sign Up"></form>

Answer 3

if($link->connect_error){

die ("connection failed: " . $link->connect_error);

}

$fname = $_POST['fname'];

$lname = $_POST['lname'];

$email = $_POST['email'];

$pno = $_POST['pno'];

$pass = $_POST['pass'];

$sql ="INSERT INTO user (firstname, lastname, email, phn_number, password)     VALUES ('$fname', '$lname', '$email', '$pno', '$pass')";

$res = mysql_query($sql);