CODEWITHSUNDEEP
phphtml
Guitarpunkmasta
Guitarpunkmasta

Reputation: 23

displaying a form with $_POST?

i have been trying to get my filled out form to display next to it. I have tried doing this with an if statement but no luck.

I also trying just doing print_r and that did work but i want it to be after u hit submit it shows.

heres my code so far!

<form action="phptest1.php" method="POST">
<fieldset>
    <legend>Inloggegevens:</legend>
    <label for="naam">Gebruikersnaam:</label>
    <input type="text" name="naam" id="naam">
    <label for="wachtwoord">Wachtwoord:</label>
    <input type="password" name="wachtwoord" id="wachtwoord"><br>
    <label for="man">Man</label><br>
    <input type="radio" name="gender" value="male"><br>
    <label for="for">Vrouw</label><br>
    <input type="radio" name="gender" value="female"><br>
    <label for="checkbox">Ik heb de algemene voorwaarden gelezen.</label><br>
    <input type="checkbox" name="conditions" value="agree">
    <textarea name="commentaar" rows="4" cols="40"
    placeholder="Schrijf hier uw commentaar…"></textarea>
    <select name="land">
        <option value="nl">Nederland</option>
        <option value="be">Belgi&euml;</option>
        <option value="de">Duitsland</option>
        <option value="fr">Frankrijk</option>
    </select>
</fieldset>

    <input type="reset"><br>
    <input for="submit" type="submit">


<?php
if (isset($_POST['submit']))
{
    echo '<h1>'.$_POST['naam'].'</h1>';
    echo '<br><br>';
    echo $_POST['wachtwoord'];
    echo '<br><br>';
    echo $_POST["gender"];
    echo '<br><br>';
    echo $_POST["land"];
    echo '<br><br>';

    } else {
    echo 'U bent nog niet ingelogd.';
    }

    ?>

Upvotes: 0

Views: 50

Answers (3)

AraByte
AraByte

Reputation: 154

Replace your submit button with the following :

<input type="submit" name="submit" value="submit"/>

And make sure the post is submitted to the same page by removing phptest1.php from action or just use:

<form action="<?php echo $_SERVER["PHP_SELF"];?>" method="post">

Upvotes: 0

user3783243
user3783243

Reputation: 5224

The indices of $_POST are populated by the name of the input elements. Since you have no element named submit your isset is never met. To fix that give it a name.

<input for="submit" type="submit" name="submit">

Upvotes: 2

SuperDJ
SuperDJ

Reputation: 7661

Instead of 

action="phptest.php"

if(isset($_POST['submit']))

You can just use 

action=""

if($_POST)

The action="" will make sure the post is send to the current page/ file again. The if($_POST) still makes sure that the code inside the if statement is only run when the page is posted. You could use isset($_POST['submit']) however your submit button doesn't have a name of submit so there will never be a $_POST['submit'].

Upvotes: 0

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