Question

This can be done in Javascript with isNAN instead of !isset.
Using the example below – both forms post to my script, one without a value and one with a value. Is the below code a correct way to do this in PHP to assign a value if the post var is not present?

$mycheck = !isset($_POST[‘value’]) ? 0 : $_POST[‘value’];

<? 
if($_POST) :
  $mycheck = !isset($_POST['value']) ? 0 : $_POST['value'];
  echo $mycheck;
endif;
?>

<!-- send value-->
<form action="" method="post">
  <select name="value">
    <option value="0">0</option>
    <option value="5">5</option>
  </select>
  <input type="submit" name="submit">
</form>

<!-- doesn't send value-->
<form action="" method="post">
  <input name="different_var">
  <input type="submit" name="submit">
</form>

Answer 1

Yes, that’s an acceptable way of setting default values.

An alternative, introduced in PHP 5.3, is to omit the middle argument of your ternary expression:

$mycheck = $_POST['value'] ?: 0;

However, this will likely throw notices for trying to access an array with a non-existant key if $_POST['value'] isn’t set. Therefore your original method is more-accepted.