#51013 [Bgs]: use namespace in eval
| From: | dellytufy at hotmail dot com | Date: | Thu, 11 Feb 2010 13:21:19 +0000 |
| Subject: | #51013 [Bgs]: use namespace in eval | ||
| References: | 1 | Groups: | php.bugs |
| Request: | Send a blank email to php-bugs+get-147436@lists.php.net to get a copy of this message | ||
ID: 51013
User updated by: dellytufy at hotmail dot com
Reported By: dellytufy at hotmail dot com
Status: Bogus
Bug Type: Unknown/Other Function
Operating System: win
PHP Version: 5.3SVN-2010-02-11 (SVN)
New Comment:
thanx i will check again
but
why it's running
eval('use \d3\sql\\'.$type.' as baglanti; $a=new baglanti();');
it's not
eval('use \d3\sql\\'.$type.' as baglanti;');
Previous Comments:
------------------------------------------------------------------------
[2010-02-11 12:11:43] johannes@php.net
Thank you for taking the time to write to us, but this is not
a bug. Please double-check the documentation available at
http://www.php.net/manual/ and the instructions on how to
report
a bug at http://bugs.php.net/how-to-report.php
The use in the eval() is valid for the eval\'d code.
------------------------------------------------------------------------
[2010-02-11 11:53:49] dellytufy at hotmail dot com
Description:
------------
i cannot use namespace in eval()
as
eval('use namespaceName\classname as myClass;');
after this i cannot ust $mc=myClass(); undeclared class
namespace d3;
namespace d3\sql;
class oracle{
public $type='oracle';
}
namespace d3\linq;
class linq{
public $className='linq';
}
$type='oracle';
eval('use \d3\sql\\'.$type.' as baglanti;');
$a=new baglanti();// error. undeclared class d3\linq\baglanti
echo $a->type;
Reproduce code:
---------------
$type='oracle';
eval('use \d3\sql\\'.$type.' as baglanti; $a=new baglanti();');
echo $a->type;
but this is successfully runned.
Expected result:
----------------
namespace d3;
namespace d3\sql;
class oracle{
public $type='oracle';
}
namespace d3\linq;
class linq{
public $className='linq';
}
$type='oracle';
eval('use \d3\sql\\'.$type.' as baglanti;');
$a=new baglanti();
echo $a->type;
i think it must write oracle .
------------------------------------------------------------------------
--
Edit this bug report at http://bugs.php.net/?id=51013&edit=1