Re: [PHP] Purpose of $$var ?????
Variable variable. Read the docs.
$v = 'foo';
$foo = 'bar';
echo $$v;
Regards,
Andrey
P.S.
Sometimes {} are used : ${$v}
Scott Fletcher [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]...
The script was working great before PHP 4.2.x and not after that. So, I
looked through the code and came upon this variable, $$var. I have no
idea what the purpose of the double $ is for a variable. Anyone know?
--clip--
$var = v.$counter._high_indiv;
$val3 = $$var;
--clip
Thanks,
FletchSOD
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Re: [PHP] Purpose of $$var ?????
Interesting! Look like the 2nd $ is decomissioned and is reserve for
something in the future or something. Just like the _ is when it come
with $_POST as an example. That would explain why it doesn't work with PHP
4.2.x up.
Andrey Hristov [EMAIL PROTECTED] wrote in message
002601c22cd0$b1995170$1601a8c0@nik">news:002601c22cd0$b1995170$1601a8c0@nik...
Variable variable. Read the docs.
$v = 'foo';
$foo = 'bar';
echo $$v;
Regards,
Andrey
P.S.
Sometimes {} are used : ${$v}
Scott Fletcher [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]...
The script was working great before PHP 4.2.x and not after that. So, I
looked through the code and came upon this variable, $$var. I have no
idea what the purpose of the double $ is for a variable. Anyone know?
--clip--
$var = v.$counter._high_indiv;
$val3 = $$var;
--clip
Thanks,
FletchSOD
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RE: [PHP] Purpose of $$var ?????
The cause for your problem would be that register_globals defaults to off in PHP 4.2.x and greater. The solution? Start using the new superglobals ($_POST, $_GET, $_SESSION etc) or (not recomended) set register_globals = on in php.ini Read more here: http://www.php.net/manual/en/language.variables.predefined.php Regards Joakim Andersson -Original Message- From: Scott Fletcher [mailto:[EMAIL PROTECTED]] Sent: Tuesday, July 16, 2002 3:54 PM To: [EMAIL PROTECTED] Subject: [PHP] Purpose of $$var ? The script was working great before PHP 4.2.x and not after that. So, I looked through the code and came upon this variable, $$var. I have no idea what the purpose of the double $ is for a variable. Anyone know? --clip-- $var = v.$counter._high_indiv; $val3 = $$var; --clip Thanks, FletchSOD -- 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] Purpose of $$var ?????
variable variable... right up there with array array
basically what you are saying is resolve $var, then find out what that variable
holds
example;
assume your $counter is currently at 5
$var = v.$counter._high_indiv;
would mean that $var= v5_high_indiv
assuming that v5_high_indiv is dynamically assigned somewhere as a variable
$$var is the value of $v5_high_indiv
make sense?
variable variables are especially good in loops... for example, if you have
variables called
$user1, $user2, $user3
to print out all the variables would require one line per variable (and alot of
typing). using variable variables you could print out the value of all users by
looping it
for($i=1;$i100;$i++){
$user=user.$i;
echo $$user;
}
Dave
-Original Message-
From: Scott Fletcher [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, July 16, 2002 9:54 AM
To: [EMAIL PROTECTED]
Subject: [PHP] Purpose of $$var ?
The script was working great before PHP 4.2.x and not after that. So, I
looked through the code and came upon this variable, $$var. I have no
idea what the purpose of the double $ is for a variable. Anyone know?
--clip--
$var = v.$counter._high_indiv;
$val3 = $$var;
--clip
Thanks,
FletchSOD
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Re: [PHP] Purpose of $$var ?????
I tried that test script you mentioned and it doesn't work in PHP 4.2.1. I
have a very good idea why is that, must have to do with the php.ini.
Unfortunately, it doesn't work either. I'll tell you what, I'll just throw
out that script and write a different script. This time, no double $.
--clip--
for($i=1;$i100;$i++){
$user=user.$i;
echo $$user.*br;
}
--clip--
Joakim Andersson [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
The cause for your problem would be that register_globals defaults to off
in
PHP 4.2.x and greater.
The solution? Start using the new superglobals ($_POST, $_GET, $_SESSION
etc) or (not recomended) set register_globals = on in php.ini
Read more here:
http://www.php.net/manual/en/language.variables.predefined.php
Regards
Joakim Andersson
-Original Message-
From: Scott Fletcher [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, July 16, 2002 3:54 PM
To: [EMAIL PROTECTED]
Subject: [PHP] Purpose of $$var ?
The script was working great before PHP 4.2.x and not after
that. So, I
looked through the code and came upon this variable, $$var.
I have no
idea what the purpose of the double $ is for a variable.
Anyone know?
--clip--
$var = v.$counter._high_indiv;
$val3 = $$var;
--clip
Thanks,
FletchSOD
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Re: [PHP] Purpose of $$var ?????
What docs at php.net? under variable, predefine variable or what?
Andrey Hristov [EMAIL PROTECTED] wrote in message
002601c22cd0$b1995170$1601a8c0@nik">news:002601c22cd0$b1995170$1601a8c0@nik...
Variable variable. Read the docs.
$v = 'foo';
$foo = 'bar';
echo $$v;
Regards,
Andrey
P.S.
Sometimes {} are used : ${$v}
Scott Fletcher [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]...
The script was working great before PHP 4.2.x and not after that. So, I
looked through the code and came upon this variable, $$var. I have no
idea what the purpose of the double $ is for a variable. Anyone know?
--clip--
$var = v.$counter._high_indiv;
$val3 = $$var;
--clip
Thanks,
FletchSOD
--
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To unsubscribe, visit: http://www.php.net/unsub.php
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Re: [PHP] Purpose of $$var ?????
Alright! Found the problem! Faulty script written that come before this
script where $$var come into play. At least, it wasn't me, it was the other
programmer's error. :-)
Scott Fletcher [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Interesting! Look like the 2nd $ is decomissioned and is reserve for
something in the future or something. Just like the _ is when it come
with $_POST as an example. That would explain why it doesn't work with
PHP
4.2.x up.
Andrey Hristov [EMAIL PROTECTED] wrote in message
002601c22cd0$b1995170$1601a8c0@nik">news:002601c22cd0$b1995170$1601a8c0@nik...
Variable variable. Read the docs.
$v = 'foo';
$foo = 'bar';
echo $$v;
Regards,
Andrey
P.S.
Sometimes {} are used : ${$v}
Scott Fletcher [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]...
The script was working great before PHP 4.2.x and not after that. So,
I
looked through the code and came upon this variable, $$var. I have
no
idea what the purpose of the double $ is for a variable. Anyone
know?
--clip--
$var = v.$counter._high_indiv;
$val3 = $$var;
--clip
Thanks,
FletchSOD
--
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