Help with Arrays.

Hi! I'm kinda new to php and I am stumped by a dilemma.

Okay I have a string called $word which contains "global" or any other
word.
Then I also have a set or arrays which is named after the strings in
the $word variable like $global[]
I also have a value called $tem5 depicting the current array index.

A short version would be

$global[1]="some string";
$word="global";
$temp5=1;

How can I get the value of $global[1] then using the $word and $temp
values?

I tried
$wordn = $word[$temp5]; <==obviously didn't work;

parse_str("" . "worrn=(" . $word. "[" . $temp5 ."])"); <==somehow
didn't work
parse_str("" . "worrn=($" . $word. "[" . $temp5 ."])"); <==somehow
didn't work
parse_str("" . "worrn=(\$" . $word. "[" . $temp5 ."])"); <==somehow
didn't work

Then I tried to check if i can actually pass on the values
$wordn = $global[1]; <==which worked.

So how can I access an array if I have its name and index?
Please help.

fogger schreef:

> A short version would be
>
> $global[1]="some string";
> $word="global";
> $temp5=1;
>
> How can I get the value of $global[1] then using the $word and $temp
> values?
>

$$word[$temp5];

Note the double $ before 'word'.


JW

Janwillem Borleffs wrote:

> fogger schreef:
>
>> A short version would be
>>
>> $global[1]="some string";
>> $word="global";
>> $temp5=1;
>>
>> How can I get the value of $global[1] then using the $word and $temp
>> values?
>>
>
> $$word[$temp5];
>
> Note the double $ before 'word'.
>
>
> JW

more correct way:
$wordn = ${$word}[$temp5];

--
a.d.

Andrius Dijakas schreef:
>
> more correct way:
> $wordn = ${$word}[$temp5];
>

Doesn't make a difference, really.


JW

Janwillem Borleffs wrote:

> Andrius Dijakas schreef:
>>
>> more correct way:
>> $wordn = ${$word}[$temp5];
>>
>
> Doesn't make a difference, really.
>
>
> JW

but on my PHP v5.2.6 server I get "Notice: Undefined variable" with
$$word[$temp5];
while no problem with
${$word}[$temp5];

--
a.d.

Andrius Dijakas schreef:
> but on my PHP v5.2.6 server I get "Notice: Undefined variable" with
> $$word[$temp5];
> while no problem with
> ${$word}[$temp5];

Indeed, I get the same problem with PHP4... The $$name notation of
variable variables only seems to be valid for scalars.


JW

On Thu, 08 May 2008 10:08:24 +0200, Janwillem Borleffs <jw@jwscripts.com>
wrote:

> Andrius Dijakas schreef:
>> but on my PHP v5.2.6 server I get "Notice: Undefined variable" with
>> $$word[$temp5];
>> while no problem with ${$word}[$temp5];
>
> Indeed, I get the same problem with PHP4... The $$name notation of
> variable variables only seems to be valid for scalars.

The problem is just that $word[$temp5] DOES exist, it means $word[1], and
if $word = 'global', $word[1] = 'l' (lower case L for those that get
confused with 1)), and $l obvisouly doesn't exist. The curly braces will
tell PHP what portion should be the variable name, else it probably
assumes ${$word[$temp5]}.

A far less error-prone solution for the OP would obviously be:

<?php
$data = array();
$data['global'][1]="some string";
$word="global";
$temp5=1;
echo $data[$word][$temp5];
?>
--
Rik Wasmus

Thanks for the quick replies!

Very much appreciated. :)

On 8 May, 08:42, Andrius Dijakas <andri...@freemail.lt> wrote:
> Janwillem Borleffs wrote:
> > Andrius Dijakas schreef:
>
> >> more correct way:
> >> $wordn = ${$word}[$temp5];
>
> > Doesn't make a difference, really.
>
> > JW
>
> but on my PHP v5.2.6 server I get "Notice: Undefined variable" with
> $$word[$temp5];
> while no problem with
> ${$word}[$temp5];
>
> --
> a.d.

I don't get an error (earlier PHP) but agree that ${$word}[$temp5] is
more readable and avoid the ambigious interpretation ${$word[$temp5]}

C.