CODEWITHSUNDEEP
phpphp-5.2
Alix Axel
Alix Axel

Reputation: 154543

Unexpected T_PAAMAYIM_NEKUDOTAYIM in PHP 5.2.x

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?

Upvotes: 6

Views: 14348

Answers (5)

T.Todua
T.Todua

Reputation: 56361

Another case :

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

Upvotes: 0

Rajan Rawal
Rajan Rawal

Reputation: 6317

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

Upvotes: 0

Chris Baker
Chris Baker

Reputation: 50592

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;
}

Upvotes: 1

webbiedave
webbiedave

Reputation: 48897

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

Upvotes: 3

John Cartwright
John Cartwright

Reputation: 5084

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.

Upvotes: 0

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