PHP overriding isset value, showing new output - php

Question

So, I have this code
<?php
if(!isset($_POST["step1_submit"]))
{
echo "Fill step1!";
step1();
}
function step1()
{
if(isset($_POST["step1_submit"]))
{
step2();
}
else
{
echo '<form action="" method="post">STEP 1: <input name="step1_input"/><input name="step1_submit" type="submit" value=">>STEP 2>>"/></form>';
}
}
function step2()
{
echo "reached step 2";
if(isset($_POST["step2_submit"]))
{
echo '<br/>'.$_POST["step2_input"];
step3();
}
else
{
echo '<form method="post">STEP 2: <input name="step2_input"/><input name="step2_submit" type="submit" value=">>STEP 3>>"/></form>';
}
}
function step3()
{
echo '<strong>WELL DONE</strong>';
}
?>
It shows the input for step1, but never gets to show step2. In my opinion, it gets stuck on calling the function step2, because while doing it, the isset($_POST(step1_submit)) changes it value to NULL.
How can I manage to make this code work? It should be like: filling step 1 input >> getting to fill step 2 input >> submit step 2 >> get to see the 'WELL DONE' echo.

Answer 1

Basically the conditional statement on top needs an else branch:
if(!isset($_POST["step1_submit"]))
{
echo "Fill step1!";
step1();
} else {
step2();
}
...
However, if you have more steps it should look like this:
switch(TRUE) {
case isset('step1_submit') :
step2();
break;
case isset('step2_submit') :
step3();
break;
...
default:
echo "Fill step1!";
step1();
}
Then change your functions to:
function step1()
{
echo '<form action="" method="post">STEP 1: <input name="step1_input"/><input name="step1_submit" type="submit" value=">>STEP 2>>"/></form>';
}
function step2()
{
echo '<form action="" method="post">STEP 2: <input name="step2_input"/><input name="step2_submit" type="submit" value=">>STEP 3>>"/></form>';
}
...
The if conditionals aren't required there.