Simple php calculator error (new to this!)

Question

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.

When you're asking other people to put in the time to help you, instead of apologizing for messy formatting or typos, take the time to actually correct these things yourself. — Casey Rule, Dec 11 '14 at 19:24
Unchecked checkboxes and radio buttons are not posted to the server, so you should check if the "type" index is even set before trying to read it out. — Cᴏʀʏ, Dec 11 '14 at 19:25
Please take the time to learn how to [interpret errors and fix your code](http://jason.pureconcepts.net/2013/05/fixing-php-errors/). — Jason McCreary, Dec 11 '14 at 19:25
@CaseyRule It's quite a long form, and I don't have internet access on the computer I have the form on, so I did it all by hand. I do not know how to not make it look messy either. — , Dec 11 '14 at 19:28
@warren I'm new to php, but not that new to html. It's correctly wrapped. — , Dec 11 '14 at 19:28

score 1 · Answer 1 · answered Dec 11 '14 at 19:26

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";
}

answered Dec 11 '14 at 19:26

Hanky Panky

46,730
8
72
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Should this be right after my variables? – Dec 11 '14 at 19:32
"Simply check if type is posted before you pick it up" Would that mean that it'd first see if a box is checked, and then run the rest/just return if one isn't? Finding it hard to make sense of it all! – Dec 11 '14 at 19:33

Jay Harris · Answer 2 · 2014-12-12T00:59:14.957

1

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

edited Dec 12 '14 at 00:59

answered Dec 11 '14 at 19:28

Jay Harris

4,201
17
21

2

`selected="selected"` – Hanky Panky Dec 11 '14 at 19:29
@Hanky웃Panky many ways to do it (`selected`, `selected='selected'`, `selected=""`), but for exactness i changed it... – Jay Harris Dec 11 '14 at 19:33
1

Ok :) Shouldn't that be `checked`, instead of `selected`, btw for radio buttons ? :P – Hanky Panky Dec 12 '14 at 00:56
@Hanky웃Panky yes lol what an egregious mistake – Jay Harris Dec 12 '14 at 00:59

Peter · Answer 3 · 2014-12-12T09:36:27.593

0

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).

edited Dec 12 '14 at 09:36

answered Dec 11 '14 at 19:23

Peter

1,658
17
23

The form can be posted without posting the `type` variable "when I don't select any radio box for math type". Also, `$_POST['method']` will be undefined. – showdev Dec 11 '14 at 19:23
Yes of course further validation still has to be done. – Peter Dec 11 '14 at 19:25
@Peter That gave me another error notice! :( – Dec 11 '14 at 19:31
@Jimmy Oops, typo, fixed it – Peter Dec 12 '14 at 09:36
Thanks Peter, I think I got it! – Dec 16 '14 at 11:24

Edgar · Answer 4 · 2014-12-11T19:38:46.617

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!";
    }
?>

Simple php calculator error (new to this!)

4 Answers4