ID: 8860
Updated by: [EMAIL PROTECTED]
Reported By: [EMAIL PROTECTED]
-Status: Open
+Status: Closed
Bug Type: Feature/Change Request
Operating System: linux
PHP Version: 4.0.4pl1
New Comment:
you need to use eval() to make something like this unambiguous.
Previous Comments:
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[2001-01-23 10:25:22] [EMAIL PROTECTED]
It is definitely not meant to work this way. 'test[2]' is
not a variable name. $test is the variable (array), and
$test[2] is second element of the array. So it belongs to
feature requests.
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[2001-01-23 10:20:59] [EMAIL PROTECTED]
ah, sorry, should've read the report more carefully.
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[2001-01-23 10:07:21] [EMAIL PROTECTED]
changed status
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[2001-01-23 10:06:24] [EMAIL PROTECTED]
ok and that isn;t working either...
althoug i liked the first more (and still think it should be that way)
so this doesn't work either:
$three = "Grrrr";
echo "Test";
$test = Array("one","two","three","four"); $var = 'test[2]'; echo
$$var; /* should echo "three" */
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[2001-01-23 09:59:43] [EMAIL PROTECTED]
no. it will echo contents of $three. if you don't have that variable in
current scope, it'll echo null, and, depending on your settings, emit a
warning.
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The remainder of the comments for this report are too long. To view
the rest of the comments, please view the bug report online at
http://bugs.php.net/8860
--
Edit this bug report at http://bugs.php.net/?id=8860&edit=1
