RE: [PHP] Varible Varibles
I think you're missing the point of variable variables. ? $a = 'foo'; $$a = 'bar'; echo $a $$a; ? After the first use of $$a, you now have a variable called $foo with a value of 'bar'. So your echo would be echo $a $foo; I kind of consider variable variables the poor mans array. Most any solution you think of with variable variables could be better solved by using arrays. ---John Holmes... -Original Message- From: Peter [mailto:[EMAIL PROTECTED]] Sent: Wednesday, June 12, 2002 10:42 PM To: Php Subject: [PHP] Varible Varibles howdy, I'm just curious here about varible varibles ... I know that you can, well it's documented that you can, do the following ? $a = foo; $$a = bar; echo $a $$a; ? which will produce foo bar now what I am curious about is, how much of a difference does that really make when you compare it to.. ? $a = foo; $a .= bar; echo $a; ? Cheers Peter the only dumb question is the one that wasn't asked -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Varible Varibles
I'm trying to use variable variables to work on arrays:
$forest = array(a, b, c, ...);
$layer[$l]= forest;
Now I want to access all array members of $forest using $$layer:
e.g.
for($c = 0; $c $$layer[$l]; $l++) {
echo $$layer[$l][$c];
}
But this doesn't work, gives syntax error,
So my workaround is:
$new = $$layer[$l];
$new[$c] refers all elements of array $forest
Is this the best workaround or am I missing something?
I think you're missing the point of variable variables.
?
$a = 'foo';
$$a = 'bar';
echo $a $$a;
?
After the first use of $$a, you now have a variable called $foo with a
value of 'bar'.
So your echo would be echo $a $foo;
I kind of consider variable variables the poor mans array. Most any
solution you think of with variable variables could be better solved by
using arrays.
---John Holmes...
-Original Message-
From: Peter [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, June 12, 2002 10:42 PM
To: Php
Subject: [PHP] Varible Varibles
howdy,
I'm just curious here about varible varibles ... I know that you can,
well
it's documented that you can, do the following
?
$a = foo;
$$a = bar;
echo $a $$a;
?
which will produce foo bar
now what I am curious about is, how much of a difference does that
really
make when you compare it to..
?
$a = foo;
$a .= bar;
echo $a;
?
Cheers
Peter
the only dumb question is the one that wasn't asked
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Re: [PHP] Varible Varibles
On Friday 14 June 2002 00:38, John Holmes wrote: I think you're missing the point of variable variables. Quite :-) I kind of consider variable variables the poor mans array. Most any solution you think of with variable variables could be better solved by using arrays. Actually variable variables are extremely powerful in the cases where they _really_ are needed. However some people mistakenly use variable variables when they really should be using arrays. -- Jason Wong - Gremlins Associates - www.gremlins.com.hk Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * /* If I can have honesty, it's easier to overlook mistakes. -- Kirk, Space Seed, stardate 3141.9 */ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Varible Varibles
well, the first method is the same as saying $a = foo; $foo = bar; echo $a $foo; whereas the second method is appending bar to $a (thus making it foobar) In first method, you get two variables, the second, just one -Original Message- From: Peter [mailto:[EMAIL PROTECTED]] Sent: Thursday, June 13, 2002 12:42 PM To: Php Subject: [PHP] Varible Varibles howdy, I'm just curious here about varible varibles ... I know that you can, well it's documented that you can, do the following ? $a = foo; $$a = bar; echo $a $$a; ? which will produce foo bar now what I am curious about is, how much of a difference does that really make when you compare it to.. ? $a = foo; $a .= bar; echo $a; ? Cheers Peter the only dumb question is the one that wasn't asked -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] Varible Varibles
Ok,...
what sort situations would you benifit using the first method?
ie
is it more suiteable to making some conditions
eg
$a = foo;
while {$$a = bar)
echo $a;
endwhile;
-Original Message-
From: Martin Towell [mailto:[EMAIL PROTECTED]]
Sent: Thursday, 13 June 2002 12:44 PM
To: 'Peter'; Php
Subject: RE: [PHP] Varible Varibles
well, the first method is the same as saying
$a = foo;
$foo = bar;
echo $a $foo;
whereas the second method is appending bar to $a (thus making it foobar)
In first method, you get two variables, the second, just one
-Original Message-
From: Peter [mailto:[EMAIL PROTECTED]]
Sent: Thursday, June 13, 2002 12:42 PM
To: Php
Subject: [PHP] Varible Varibles
howdy,
I'm just curious here about varible varibles ... I know that you can, well
it's documented that you can, do the following
?
$a = foo;
$$a = bar;
echo $a $$a;
?
which will produce foo bar
now what I am curious about is, how much of a difference does that really
make when you compare it to..
?
$a = foo;
$a .= bar;
echo $a;
?
Cheers
Peter
the only dumb question is the one that wasn't asked
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