Unexpected T_PAAMAYIM_NEKUDOTAYIM in PHP 5.2.x

Question

I'm having a hard time understanding why I'm getting an Unexpected T_PAAMAYIM_NEKUDOTAYIM error in the following code, which seems perfecly valid to me...

class xpto
{
    public static $id = null;

    public function __construct()
    {
    }

    public static function getMyID()
    {
        return self::$id;
    }
}

function instance($xpto = null)
{
    static $result = null;

    if (is_null($result) === true)
    {
        $result = new xpto();
    }

    if (is_object($result) === true)
    {
        $result::$id = strval($xpto);
    }

    return $result;
}

Output in PHP 5.3+:

echo var_dump(instance()->getMyID()) . "\n"; // null
echo var_dump(instance('dev')->getMyID()) . "\n"; // dev
echo var_dump(instance('prod')->getMyID()) . "\n"; // prod
echo var_dump(instance()->getMyID()) . "\n"; // null

In prior versions however, I can't do $result::$id = strval($xpto);, does anyone know why?

Are there any workarounds for this problem?

Milore

I copy/pasted your code to a php file, and it runs without errors. – Tautog

@Chris: What PHP version are you using? Also, check this link: codepad.org/wpT0g3VH. – Milore

Sorry, misinterpreted your question. Posted answer below. – Tautog

Possible duplicate of PHP expects T_PAAMAYIM_NEKUDOTAYIM? – Farlie

Answer 1

After looking at codepad:

if (is_object($result) === true)
{
    $result::id = strval($xpto);
}

... should be

if (is_object($result) === true)
{
    $result::$id = strval($xpto);
}

I corrected this in a new paste, and the error still exists... just letting you know about the problem in the demo code.

EDIT

Per PHP documentation page on static keyword,

As of PHP 5.3.0, it's possible to reference the class using a variable. The variable's value can not be a keyword (e.g. self, parent and static).

Unfortunately, no detail is given as to WHY to was otherwise in prior versions, nor do I see a workaround presented in the comments.

Because the class is static, though, you should be able to change the property directly:

function instance($xpto = null)
{
    static $result = null;

    if (is_null($result) === true)
    {
        $result = new xpto();
    }

    if (is_object($result) === true)
    {
        xpto::$id = strval($xpto)
    }

    return $result;
}

Tautog

Oops, that was a typo! Still getting the error though (codepad.org/wpT0g3VH). – Milore

Interestingly, my dev server (php 5.2.7) calls the original error a "Parse error: parse error" rather than trotting out the old "T_PAAMAYIM_NEKUDOTAYIM" chestnut. – Tautog

Actually, you gave me a very good tip: codepad.org/Y9ntbREa. Thank you! =) – Milore

Answer 2

The reason for the error is simply that the syntax isn't supported in < 5.3.

However, if you're trying to just access the static variable $id, then the syntax would be:

$result::id

If you do need to access a static variable variable, then a workaround is to use reflection:

$class = new ReflectionClass($xpto);
echo $class->setStaticPropertyValue ('id', strval($xpto));

ReflectionClass

Burma

Auch, that is a PITA... $id in this case is not a variable variable. Do you know where I can read more about the support of this syntax? I've noticed that non-static properties also don't work (codepad.org/Fb1mQOvx), is this also because the syntax not being supported? – Milore

You're getting the new error because you're using $this in a static function. php.net/manual/en/language.oop5.static.php – Burma

Oh yes, forgot about that! Thank you. – Milore

Answer 3

PHP Version 5.3.3, I am not getting any errors on that code.

Output:

string(0) "" string(3) "dev" string(4) "prod" string(0) ""

Your error likely lies elsewhere. Please double check the reported line numbers.

Reedbuck

PHP 5.3 works fine as I've stated in my answer, the problem only exists on prior versions of PHP. Check the Codepad link for an example. – Milore

My apologies, I thought you said it worked in prior versions, but in 5.3 it was failing. – Reedbuck

No problem, I should have made my question more clear. Thank you. – Milore

Answer 4

I came here by the reference: Syntax error in PHP 5.2 where Chandresh mentioned your link: how ever one work around for PHP 5.2 is:

class Sample{
    public static $name;

    public function __construct(){
        self::$name = "User 1";
    }
}

$sample = new Sample();
$class = 'Sample';
$name = 'name';
$val_name = "";
$str = '$class::$$name';
eval("\$val_name = \"$str\";");
//echo $val_name."<br>";
eval("\$name = $val_name;");
echo $name;

Ignore if you already resolved. Thank you

Beta

Answer 5

Another case :

It may happen on some servers (PHP VERSION ??) if you use: if (empty(NAME_OF_A_CONSTANT)) ...

Casto