Proper way to declare multiple vars in PHP

I've been coding personal scripts for years in PHP and get used to turn off Error display. I'm about to release some of these scripts and would like to do it the proper way.

The only reason why I turn off error display is to avoid having to test every single var, prior using it, thanks to isset().

So, here is my question:

Is there a better way to declare multiple vars than this ?

<?php
  // at the begining of my main file
  if (!isset($foo))  ($foo = '');
  if (!isset($bar))  ($bar = '');
  if (!isset($ping)) ($ping = '');
  if (!isset($pong)) ($pong = '');
  // etc. for every single var
?>

Something like this for instance :

<?php
  var $foo, $bar, $ping, $pong;
?>

<?php
  $foo = $bar = $ping = $pong = '';
?>

If it's your script and you know which variables where you use, why you want spend recourses to check if the variable was declared before?

I posted this in a comment earlier, but someone suggested I submit it as an answer.

The shortest and simplest way I can think of, is to do:

$foo = $bar = $ping = $pong = '';

I often prefer to set things to false, instead of an empty string, so that you can always do checks in the future with === false, but that is just a preference and depends on how you are using these variables and for what.

Your if() with isset() attempt is the proper way of doing that!

But you can write it a little bit shorter/more readable, using the Ternary Operator:

$foo = isset($foo) ? $foo : '';

The first $foo will be set to the value after the ? when the condition after the = is true, else it will be set to the value after the :. The condition (between = and ? ) will always be casted as boolean.

Since PHP 5.3 you can write it even shorter:

$foo = isset($foo) ?: '';

This will set $foo to TRUE or FALSE (depending on what isset() returns), as pointed by @Decent Dabbler in the comments. Removing isset() will set it to '' but it will also throw an undefined variable notice (not in production though).

Since PHP 7 you can use a null coalesce operator:

$foo = $foo ?? '';

This won't throw any error, but it will evaluate as TRUE if $foo exists and is empty, as opposed to the shorthand ternary operator, that will evaluate as FALSE if the variable is empty.

A somewhat round-about way of doing this is if you put the name of your variables in an array, and then loop them with a Ternary Operator, similar to powtac's answer.

$vars = array('foo', 'bar', 'ping', 'pong');
$defaultVar = '';

foreach($vars as $var)
{
    $$var = isset($$var) ? $$var : $defaultVar;
}

As mentioned in other answers, since version 5.3, PHP allows you to write the above code as follows:

$vars = array('foo', 'bar', 'ping', 'pong');
$defaultVar = '';

foreach($vars as $var)
{
    $$var = isset($$var) ?: $defaultVar;
}

Note the changed Ternary Operator.

In OOP you can use this approach:

protected $password, $full_name, $email;

For non-OOP you declare them just in code they will be Undefined if you didn't assign any value to them:

$foo; $bar; $baz;

$set_foo =  (isset($foo)) ? $foo : $foo = 'Foo';
echo $set_foo;

Why not just set them?

<?php
$foo = '';
$bar = '';
//etc
?>

If you're trying to preserve the value in them, then yes that's the correct way in general. Note that you don't need the second pair of brackets in your statements:

if (!isset($foo)) $foo = '';

is enough.

To fix the issue of

  <?php
  $foo = $bar = $ping = $pong = '';
  ?>

throwing >Notice: Undefined variable: ...

  <?php
  @$foo = $bar = $ping = $pong = '';
  ?>

It will not fix it but it will not be shown nd will not stop the script from parsing.