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?

I copy/pasted your code to a php file, and it runs without errors. — Chris Baker, Feb 14 '11 at 18:05
@Chris: What PHP version are you using? Also, check this link: http://codepad.org/wpT0g3VH. — Alix Axel, Feb 14 '11 at 18:09
Possible duplicate of [PHP expects T\_PAAMAYIM\_NEKUDOTAYIM?](http://stackoverflow.com/questions/592322/php-expects-t-paamayim-nekudotayim) — John Slegers, Feb 24 '17 at 13:20

webbiedave · Answer 1 · 2011-02-14T18:30:34.547

3

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

edited Feb 14 '11 at 18:30

answered Feb 14 '11 at 18:17

webbiedave

48,414
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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 (http://codepad.org/Fb1mQOvx), is this also because the syntax not being supported? – Alix Axel Feb 14 '11 at 18:29
You're getting the new error because you're using $this in a static function. http://php.net/manual/en/language.oop5.static.php – webbiedave Feb 14 '11 at 18:31
Oh yes, forgot about that! Thank you. – Alix Axel Feb 14 '11 at 18:35

Chris Baker · Accepted Answer · 2011-02-14T18:59:14.830

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;
}

edited Feb 14 '11 at 18:59

answered Feb 14 '11 at 18:10

Chris Baker

49,926
12
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Oops, that was a typo! Still getting the error though (http://codepad.org/wpT0g3VH). – Alix Axel Feb 14 '11 at 18:12
1

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. – Chris Baker Feb 14 '11 at 18:28
1

Actually, you gave me a very good tip: http://codepad.org/Y9ntbREa. Thank you! =) – Alix Axel Feb 14 '11 at 18:33

score 0 · Answer 3 · answered Jul 20 '15 at 11:35

0

Another case :

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

answered Jul 20 '15 at 11:35

T.Todua

53,146
19
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237

score 0 · Answer 4 · answered Feb 14 '11 at 18:05

0

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.

answered Feb 14 '11 at 18:05

John Cartwright

5,109
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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. – Alix Axel Feb 14 '11 at 18:08
My apologies, I thought you said it worked in prior versions, but in 5.3 it was failing. – John Cartwright Feb 14 '11 at 18:13
No problem, I should have made my question more clear. Thank you. – Alix Axel Feb 14 '11 at 18:34

score 0 · Answer 5 · edited May 23 '17 at 11:33

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

Unexpected T_PAAMAYIM_NEKUDOTAYIM in PHP 5.2.x

5 Answers5

Another case :

Linked