Three forms actions in one php page

Question

Assume that I have a PHP page that has three forms.

Form one, it will ask the user to enter the email address while the code is hidden.

<label for="email">Email Address:</label> <input type="text" id="email" name="email">
<input type="hidden" id="code" name="code" value="42"> <input type="submit" value="Submit1">

Form two will be showing after step 1. In this step, the user will be asked to enter the code in step 1 (which is equal to 42).

<input type="number" id="code2" name="code2"> <input type="submit" value="Submit2">

Form three, will be showing after entering corrected code in step 2.

<input type="number" id="code2" name="code2"> <input type="submit" value="Submit3">

How can I program it?

You can put the form action as the current URL, and add ifs to the other forms, based on the stage the user is on. Is that a clear enough explanation? — tola
– tola, Commented Mar 6, 2022 at 9:40
If you can please support me with the code as I struggle with it. Appreciate your assistance. — Hisham Alghamdi
– Hisham Alghamdi, Commented Mar 6, 2022 at 9:45

Ownagin · Accepted Answer · 2022-03-06 10:59:12Z

1

You could try this approach :

<?php 
function getFormContent() {
  $email = $_POST['email'] ?? null; //$email equals to null if it is undefined, to prevent errors

  if (!$email) {
    echo '<input type="text" name="email" placeholder="Email.." />';
    echo '<input type="hidden" id="code" name="code" value="42">';
  } else {
    echo '<input type="hidden" name="email" value="'.$email.'" />'; //keep email set after submitting first step
  }

  if ($email && $_POST['code']) {
    if (isset($_POST['code2']) && $_POST['code2'] == $_POST['code']) {
      //final form
    } 
    else { //second step
      echo '<input type="number" id="code2" name="code2">';
      echo '<input type="hidden" id="code" name="code" value="42">'; //we still need this in case user submitted a wrong code
    }
  }
}
?>

<form method="post">
  <?= getFormContent() ?>
  <input type="submit" />
</form>

answered Mar 6, 2022 at 10:59

Ownagin

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3 Comments

Hisham Alghamdi Over a year ago

very nice. In case I don't want to show <input type="submit" /> in the final form, how can I do it?

Ownagin Over a year ago

You could set a variable (e.g. $finalForm = true) in the final form condition, and wrap the button submit in a if (!isset($finalForm)) condition

Hisham Alghamdi Over a year ago

Would it be possible to update it on the code please. Sorry for any incontinence caused you.

tola · Accepted Answer · 2022-03-06 11:08:21Z

1

You need to state the action of the form as the name of the file. So, if the file you are working on is name example.php it should look like so:

$email = trim(get_variable_value('email'));
$code = trim(get_variable_value('code'));
$code2 = trim(get_variable_value('code2'));

//make sure to validate the input!! if it's not valid send back

<? if(!isset($email)){ //alternatively, you can add a hidden input `stage`. I just don't like it, as it can be bypassed.
?>
    <form name="transferToNewForm" action="example.php" method="post">
        <label for="email">Email Address:</label> <input type="text" id="email" name="email">
        <input type="hidden" id="code" name="code" value="42">
        <input type="submit" value="Submit1">
    </form>
<? } 
if(isset($email) && isset($code) && $code == 42){ ?>
    <form name="transferToNewerForm" action="example.php" method="post">
        <label for="email">Email Address:</label> <input type="text" id="email" name="email" value="<?= $email ?>" readonly>
        <input type="number" id="code2" name="code2">
        <input type="submit" value="Submit2">
    </form>
<? } 
if(isset($email) && isset($code2) && $code2 == 42){ ?>
    <form name="finalFrom" action="theActualPageYouWant.php" method="post">
        <label for="email">Email Address:</label> <input type="text" id="email" name="email" value="<?= $email ?>" readonly>
        <input type="number" id="code2" name="code2">
        <input type="submit" value="Submit3">
    </form>
<? } ?>

edited Mar 6, 2022 at 11:08

answered Mar 6, 2022 at 10:11

tola

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5 Comments

Hisham Alghamdi Over a year ago

( ! ) Parse error: syntax error, unexpected ';', expecting ')' in C:\wamp64\www\form2\index.php on line 4. This is in line 4 $email = trim(get_variable_value('email');

tola Over a year ago

fixed. I was missing a ) on the first 3 lines @HishamAlghamdi

Hisham Alghamdi Over a year ago

sorry, got this error: ( ! ) Parse error: syntax error, unexpected '<' in C:\wamp64\www\form2\index.php on line 11 On this line: <form name="transferToNewForm" action="example.php" method="post">

tola Over a year ago

right, I didn't include the <?php ?> tags. I'll edit in a moment @HishamAlghamdi

Hisham Alghamdi Over a year ago

I still get errors: Notice: Undefined variable: email in C:\wamp64\www\f2\index.php on line 18 Call Stack #TimeMemoryFunctionLocation 10.0022238936{main}( )...\index.php:0 " readonly> and Notice: Undefined variable: email in C:\wamp64\www\f2\index.php on line 25 Call Stack #TimeMemoryFunctionLocation 10.0022238936{main}( )...\index.php:0 " readonly>

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