PHP: $_POST cannot retrieve existing inputs - php

Question

I'm attempting to make a button that, based on the selected form/inputs on the page, will bring you to a page called "typeDefine.php?openness=3?conscientiousness=2?extroversion=1?agreeableness=2?neuroticism=1"(the numbers varying based on the selected inputs). However, $selectedNum- the variable that would ideally be containing the $_POST for each input- is returning an error immediately once the page is loaded, saying:
Undefined index
<?php
$typeWords = array("openness", "conscientiousness", "extroversion", "agreeableness", "neuroticism");
$typeLetters = array("o", "c", "e", "a", "n");
$typePath = "";
$correspondingLetters = array("I", "II", "III");
$isFirst = true;
foreach($typeWords as $typeWord)
{
$selectedNum = $_POST[$typeWord];//error here!!!
if(isset($selectedNum))//if got $typeWord in a form
{
$separationChar;
if($isFirst)
{
$separationChar = "?";
$isFirst = false;
}
else
{
$separationChar = "&";
}
$typePath = $typePath . $separationChar . $typeWord . "=" . $selectedNum;//e.g. $typePath = "?openness=3?conscientiousness=2?extroversion=1?agreeableness=2?neuroticism=1" for $_GET method after arriving on next page
}
}
echo 'search for type
<div>';
foreach($typeWords as $typeWord)
{
$typeLetter = substr($typeWord, 0, 1);
echo '<form method = "post" class = "column">';
for($i = 1; $i <= 3; $i++)
{
echo '<input type = "radio" name = "' . $typeWord . '" id = "' . $typeLetter . $i . '"><label for = "' . $typeLetter . $i . '">' . $correspondingLetters[$i - 1] . '</label>';//sets each input name to $typeWord for $_POST above
}
echo '<li class = "textHighlight">' . $typeWord . '</li>
</form>';
}
echo '</div>';
?>
What can I do to fix this error, in turn filling $typePath and making the script correctly bring you to the desired url upon the button's click?
Thanks in advance!

Answer 1

You should perform the isset() test on the $_POST element, not the variable that you set from it.
foreach ($typeWords as $typeWord)
{
if (isset($_POST[$typeWord])) {
$selectedNum = $_POST[$typeWord];
$typePath = $typePath . "?" . $typeWord . "=" . $selectedNum;
}
}
Note that multiple parameters need to be separated with &, ? should only be used at the beginning. There's a built-in function that will create a query string from an array, you can use that:
$typeArray = [];
foreach ($typeWords as $typeWord)
{
if (isset($_POST[$typeWord])) {
$selectedNum = $_POST[$typeWord];
$typeArray[$typeWord] = $selectedNum;
}
}
$typePath = $typePath . "?" . http_build_query($typeArray);
You can also replace the loop with:
$typeArray = array_intersect_key($_POST, array_flip($typeWords));

Answer 2

You execute your code immidiately without a present $_POST array.
You have to wrap your post related code with an if statement like this:
if ($_POST) {
$selectedNum = $_POST[$typeWord];
}