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Request #8860 variable array not working
Submitted: 2001-01-23 09:57 UTC Modified: 2002-04-27 14:25 UTC
From: wico at cnh dot nl Assigned:
Status: Closed Package: Feature/Change Request
PHP Version: 4.0.4pl1 OS: linux
Private report: No CVE-ID: None
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 [2001-01-23 09:57 UTC] wico at cnh dot nl 
according to Zeev this should work:

$test = Array("one","two","three","four"); $var = 'test[2]'; echo $$var; /* should echo "three" */

But with it doesn't


Greetz,

Wico

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 [2001-01-23 09:59 UTC] cynic@php.net 
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.
 [2001-01-23 10:06 UTC] wico at cnh dot nl 
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" */
 [2001-01-23 10:07 UTC] wico at cnh dot nl 
changed status
 [2001-01-23 10:20 UTC] cynic@php.net 
ah, sorry, should've read the report more carefully.
 [2001-01-23 10:25 UTC] stas@php.net 
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.
 [2002-04-27 14:25 UTC] jimw@php.net 
you need to use eval() to make something like this unambiguous.
 
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