Hello,
Using dynamic class names works fine without namespaces, but
doesn't work at all with namespaces. Take this simple example:
<?php
namespace test;
class Foo {}
$class = "Foo";
$foo = new test::$class();
?>
Will produce the following message:
Fatal error: Class 'test::test' not found in foo.php on line Y
Now if Foo would have a static method and we want to use dynamic
access for a static method, we get another error:
<?php
include 'foo.php';
$class = "Foo";
test::$class::test();
?>
Parse error: syntax error, unexpected T_PAAMAYIM_NEKUDOTAYIM in ...
All tested with current 5.3
/Magnus
--
Small things make base men proud.
-- William Shakespeare, "Henry VI"
Using dynamic class names works fine without namespaces, but
doesn't work at all with namespaces. Take this simple example:<?php
namespace test;
class Foo {}
$class = "Foo";
$foo = new test::$class();
?>Will produce the following message:
Fatal error: Class 'test::test' not found in foo.php on line Y
I don't think partially variable class name is supposed to work. It is
interesting though how the engine managed to parse this at all...
Stanislav Malyshev, Zend Software Architect
stas@zend.com http://www.zend.com/
(408)253-8829 MSN: stas@zend.com
Using dynamic class names works fine without namespaces, but
doesn't work at all with namespaces. Take this simple example:<?php
namespace test;
class Foo {}
$class = "Foo";
$foo = new test::$class();
?>Will produce the following message:
Fatal error: Class 'test::test' not found in foo.php on line YI don't think partially variable class name is supposed to work. It is
interesting though how the engine managed to parse this at all...
Ahh, I see, that's a bit disappointing. Any reason why it doesn't/shouldn't
work ?
/Magnus
--
One of the most striking differences between a cat and a lie is that a cat has
only nine lives.
-- Mark Twain, "Pudd'nhead Wilson's Calendar"
Ahh, I see, that's a bit disappointing. Any reason why it doesn't/shouldn't
work ?
Yes. Test::Foo is a class name, so you can't make part of it variable
and part not, just as you couldn't write:
$var = "bar";
$a = new foo$var();
and expect new instance of class "foobar". You'd have to compose the
full name in variable and then use it.
--
Stanislav Malyshev, Zend Software Architect
stas@zend.com http://www.zend.com/
(408)253-8829 MSN: stas@zend.com