Simple php calculator error (new to this!) [duplicate] - php

Question

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"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 8 years ago.
When I say 'new', I mean that this is just about the first time I'm attempting php.
Anyway. I keep getting this error notice "Undefined index: type in c:\x\calculator.php on line 33", but it still echoes "You forgot to pick mathtype!" and the calculator works fine. This error notice occurs only when I don't select any radio box for math type (+-/*).
//Part of the form
<form action="calculator.php" method="post">
<input type="text" name="1stnumber">
<input type="text" name="2ndnumber">
<input type="radio" name="type" value="addition">
<input type="radio" name="type" value="subtraction">
<input type="submit" name="send">
<?php
//My variables
$number = $_POST['1stnumber']
$numbero = $_POST['2ndnumber']
$mathtype = $_POST['type'] /* **<-line 33** */
//The calculation part of the form here, which is working
//Tell the user if he didn't pick a math type (+-)
if(is_null($mathtype)){
echo "You forgot to pick mathtype!"
}
?>
Tried with elseif as well.. I don't see what's wrong between line 33 and the if(is_null()) line!
Sorry if it looks poor, messy, or if something doesn't make sense. Might be a few typos as well. Any help is appreciated.

Answer 1

Simply check if type is posted before you pick it up
if(isset($_POST['type']))
{
$mathtype = $_POST['type'];
}
else
{
echo "Type was not selected";
}

Answer 2

Set a default selected option, using the checked attribute.
<label><input type="radio" name="type" value="addition" checked="checked"> +</label>
<label><input type="radio" name="type" value="subtraction"> -</label>
Don't forget inputs needs labels in html if left out you can use a placeholder attribute, but that's clearly not possible with type="radio"; therefore wrap the input in a label with the text description next to it eg + or -
Also, is this a copy and paste error, bc all php statements must be terminated with a semicolon ;
$number = $_POST['1stnumber']; // <- terminate
$numbero = $_POST['2ndnumber']; // <- terminate
$mathtype = $_POST['type']; // <- terminate
echo "You forgot to pick mathtype!"; // <- terminate

Answer 3

It's always good practice to check if the variable you're trying to retrieve from $_POST has actually been set, try this:
<?php
//My variables
if (isset($_POST['1stnumber'])) {
$number = $_POST['1stnumber'];
}
if (isset($_POST['2ndnumber'])) {
$numbero = $_POST['2ndnumber'];
}
if (isset($_POST['type'])) {
$mathtype = $_POST['type']; /* **<-line 33** */
}
//The calculation part of the form here, which is working
//Tell the user if he didn't pick a math type (+-)
if (is_null($mathtype)) {
echo "You forgot to pick mathtype!";
}
?>

Answer 4

Check if the form is posted:
<?php
if($_SERVER['REQUEST_METHOD'] == 'POST') {
//My variables
$number = $_POST['1stnumber']
$numbero = $_POST['2ndnumber']
$mathtype = $_POST['type'] /* **<-line 33** */
//The calculation part of the form here, which is working
//Tell the user if he didn't pick a math type (+-)
if(is_null($mathtype)){
echo "You forgot to pick mathtype!"
}
}
?>
Else the is_null check will also be executed on the first load (before the form has been posted).