Question

How can I check if user gave input to input-form and say if there is an error? I tried to put

if (!isset($_POST['email'])) {
  die('You have to give an email address.');
}

but the problem here is that it appears also if the user goes to the page first time and has not gave any input. XHTML is the following:

<form action=<?php echo SITEADDRESS.'register.php'?> method="post">
 <p>
  <label for="email">Sähköposti:</label>
  <input type="text" id="email" name="email" />
  <label for="password">Salasana:</label>
  <input type="password" id="pwd" name="pwd" />
 </p>
 <p>
  <input type="submit" value="Rekisteröidy" />
 </p>
</form> 

Answer 1

but the problem here is that it
appears also if the user goes to the
page first time and has not gave any
input. XHTML is the following:

You need to do so when user clicks the submit button, so your code should look like this:

if (isset($_POST['submit'])) {
  if (empty($_POST['email'])) {
    die('You have to give an email address.');
  }  
}

Now the above code will only execute when a user clicks the submit button unlike when he visits the page without clicking the button. Hence, above code will execute only after a user click the submit button.

You should give the name to your submit button:

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

Note: Rather than using die, you could show a message and continue to show your form something like this:

if (isset($_POST['submit'])) {
  if (empty($_POST['email'])) {
    echo 'You have to give an email address.';
  }  
}