How to keep radio button value inside php variable without error - php

Question

Let me explain more in details what I am trying to do.
Okay so, I am doing a huge form to send in a particular e-mail. My problem right now is that only my radio buttons in my form show an error when the form is submit and return false. Here's an example of what I am talking about :
<?php
if (isset($_POST['submit'])) {
$radio = $_POST['number'];
$name = $_POST['name'];
}
?>
<form action="<?= $_SERVER['PHP_SELF'] ?>" method="post">
<input type="radio" name="number" value="allo">
<input type="radio" name="number" value="yo">
<input type="text" name="name">
<input type="submit" name="submit">
</form>
So as you see, if the customer press on "submit" and didn't filled values inside of radio and input field, $name will show no error but $radio will show an error.
My Error :
Notice: Undefined index: number in D:\xampp\htdocs\test\preg.php on line 5

Answer 1

What you are getting is the server warning and not an actual error.
To avoid it you have flags that you can set to ignore warning, but this isn't a good practice.
I would suggest you to check if the form are set and handle them if not.
Code:
<?php
if (isset($_POST['submit'])) {
$radio = (isset($_POST['number']) ? $_POST['number'] : "");
$name = (isset($_POST['name']) ? $_POST['name'] : "");
}
?>
Edit: Of course, lets not forget, the better practice will be handle them against malicious code.

Answer 2

You get that error because $_POST['number'] is not passed to the page.
You should set at least one of the radio buttons, and if you want to be sure of not getting an error even if the users modifies the page, you can check with an isset as others as wrote in the other answers.
If you were wondering why you didn't get any error from the name, that's because it contains the empty string "".
Check this code and view the source of the page to see what print_r writes in a formatted way
<?php
if (isset($_POST['submit'])) {
$radio = $_POST['number'];
$name = $_POST['name'];
}
print_r($_POST)
?>
<form action="<?= $_SERVER['PHP_SELF'] ?>" method="post">
<input type="radio" name="number" value="allo" checked="true">
<input type="radio" name="number" value="yo">
<input type="text" name="name">
<input type="submit" name="submit">
</form>

Answer 3

You can write
if (isset($_POST['name'])) $name = $_POST['name'];
but even better is
isset($_POST['name']) ? $name = $_POST['name'] : $name = '';

Answer 4

if (isset($_POST['number']))
{
$radio = $_POST['number'];
}