Re: [PHP] Variable variables into an array.

2010-08-11 Thread Richard Quadling
On 11 August 2010 13:58, Bob McConnell  wrote:
> From: Richard Quadling
>
>> Quick set of eyes needed to see what I've done wrong...
>>
>> The following is a reduced example ...
>>
>> > $Set = array();
>> $Entry = 'Set[1]';
>            ^^
> Shouldn't that be $Set[1]?
>
>> $Value = 'Assigned';
>> $$Entry = $Value;
>> print_r($Set);
>> ?>
>
> Bob McConnell
>

No.

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RE: [PHP] Variable variables into an array.

2010-08-11 Thread Bob McConnell
From: Richard Quadling

> Quick set of eyes needed to see what I've done wrong...
> 
> The following is a reduced example ...
> 
>  $Set = array();
> $Entry = 'Set[1]';
^^
Shouldn't that be $Set[1]?

> $Value = 'Assigned';
> $$Entry = $Value;
> print_r($Set);
> ?>

Bob McConnell

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Re: [PHP] Variable variables into an array.

2010-08-11 Thread Richard Quadling
On 10 August 2010 18:08, Andrew Ballard  wrote:
> On Tue, Aug 10, 2010 at 12:23 PM, Richard Quadling  
> wrote:
>> On 10 August 2010 16:49, Jim Lucas  wrote:
>>> Richard Quadling wrote:

 Hi.

 Quick set of eyes needed to see what I've done wrong...

 The following is a reduced example ...

 >>> $Set = array();
 $Entry = 'Set[1]';
 $Value = 'Assigned';
 $$Entry = $Value;
 print_r($Set);
 ?>

 The output is an empty array.

 Examining $GLOBALS, I end up with an entries ...

    [Set] => Array
        (
        )

    [Entry] => Set[1]
    [Value] => Assigned
    [Set[1]] => Assigned


 According to http://docs.php.net/manual/en/language.variables.basics.php,
 a variable named Set[1] is not a valid variable name. The [ and ] are
 not part of the set of valid characters.

 In testing all the working V4 and V5 releases I have, the output is
 always an empty array, so it looks like it is me, but the invalid
 variable name is an issue I think.

 Regards,

 Richard.

 NOTE: The above is a simple test. I'm trying to map in nested data to
 over 10 levels.
>>>
>>> For something like this, a string that looks like a nested array reference,
>>> you might need to involve eval for it to "derive" that nested array.
>>>
>>
>> I'm happy with that.
>>
>> It seems variable variables can produce variables that do not follow
>> the same naming limitations as normal variables.
>>
>
> It would seem so. If eval() works, can you rearrange the strings a
> little to make use of parse_str() and avoid the use of eval()?
>
> Andrew
>

php -r "parse_str('a[1][2][3]=richard quadling'); var_dump($a);"

outputs ...

array(1) {
  [1]=>
  array(1) {
[2]=>
array(1) {
  [3]=>
  string(16) "richard quadling"
}
  }
}

Perfect.

Thanks.

-- 
Richard Quadling.

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Re: [PHP] Variable variables into an array.

2010-08-10 Thread Andrew Ballard
On Tue, Aug 10, 2010 at 12:23 PM, Richard Quadling  wrote:
> On 10 August 2010 16:49, Jim Lucas  wrote:
>> Richard Quadling wrote:
>>>
>>> Hi.
>>>
>>> Quick set of eyes needed to see what I've done wrong...
>>>
>>> The following is a reduced example ...
>>>
>>> >> $Set = array();
>>> $Entry = 'Set[1]';
>>> $Value = 'Assigned';
>>> $$Entry = $Value;
>>> print_r($Set);
>>> ?>
>>>
>>> The output is an empty array.
>>>
>>> Examining $GLOBALS, I end up with an entries ...
>>>
>>>    [Set] => Array
>>>        (
>>>        )
>>>
>>>    [Entry] => Set[1]
>>>    [Value] => Assigned
>>>    [Set[1]] => Assigned
>>>
>>>
>>> According to http://docs.php.net/manual/en/language.variables.basics.php,
>>> a variable named Set[1] is not a valid variable name. The [ and ] are
>>> not part of the set of valid characters.
>>>
>>> In testing all the working V4 and V5 releases I have, the output is
>>> always an empty array, so it looks like it is me, but the invalid
>>> variable name is an issue I think.
>>>
>>> Regards,
>>>
>>> Richard.
>>>
>>> NOTE: The above is a simple test. I'm trying to map in nested data to
>>> over 10 levels.
>>
>> For something like this, a string that looks like a nested array reference,
>> you might need to involve eval for it to "derive" that nested array.
>>
>
> I'm happy with that.
>
> It seems variable variables can produce variables that do not follow
> the same naming limitations as normal variables.
>

It would seem so. If eval() works, can you rearrange the strings a
little to make use of parse_str() and avoid the use of eval()?

Andrew

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Re: [PHP] Variable variables into an array.

2010-08-10 Thread Richard Quadling
On 10 August 2010 16:49, Jim Lucas  wrote:
> Richard Quadling wrote:
>>
>> Hi.
>>
>> Quick set of eyes needed to see what I've done wrong...
>>
>> The following is a reduced example ...
>>
>> > $Set = array();
>> $Entry = 'Set[1]';
>> $Value = 'Assigned';
>> $$Entry = $Value;
>> print_r($Set);
>> ?>
>>
>> The output is an empty array.
>>
>> Examining $GLOBALS, I end up with an entries ...
>>
>>    [Set] => Array
>>        (
>>        )
>>
>>    [Entry] => Set[1]
>>    [Value] => Assigned
>>    [Set[1]] => Assigned
>>
>>
>> According to http://docs.php.net/manual/en/language.variables.basics.php,
>> a variable named Set[1] is not a valid variable name. The [ and ] are
>> not part of the set of valid characters.
>>
>> In testing all the working V4 and V5 releases I have, the output is
>> always an empty array, so it looks like it is me, but the invalid
>> variable name is an issue I think.
>>
>> Regards,
>>
>> Richard.
>>
>> NOTE: The above is a simple test. I'm trying to map in nested data to
>> over 10 levels.
>
> For something like this, a string that looks like a nested array reference,
> you might need to involve eval for it to "derive" that nested array.
>

I'm happy with that.

It seems variable variables can produce variables that do not follow
the same naming limitations as normal variables.



-- 
Richard Quadling.

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Re: [PHP] Variable variables into an array.

2010-08-10 Thread Jim Lucas

Richard Quadling wrote:

Hi.

Quick set of eyes needed to see what I've done wrong...

The following is a reduced example ...



The output is an empty array.

Examining $GLOBALS, I end up with an entries ...

[Set] => Array
(
)

[Entry] => Set[1]
[Value] => Assigned
[Set[1]] => Assigned


According to http://docs.php.net/manual/en/language.variables.basics.php,
a variable named Set[1] is not a valid variable name. The [ and ] are
not part of the set of valid characters.

In testing all the working V4 and V5 releases I have, the output is
always an empty array, so it looks like it is me, but the invalid
variable name is an issue I think.

Regards,

Richard.

NOTE: The above is a simple test. I'm trying to map in nested data to
over 10 levels.


For something like this, a string that looks like a nested array 
reference, you might need to involve eval for it to "derive" that nested 
array.


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[PHP] Variable variables into an array.

2010-08-10 Thread Richard Quadling
Hi.

Quick set of eyes needed to see what I've done wrong...

The following is a reduced example ...



The output is an empty array.

Examining $GLOBALS, I end up with an entries ...

[Set] => Array
(
)

[Entry] => Set[1]
[Value] => Assigned
[Set[1]] => Assigned


According to http://docs.php.net/manual/en/language.variables.basics.php,
a variable named Set[1] is not a valid variable name. The [ and ] are
not part of the set of valid characters.

In testing all the working V4 and V5 releases I have, the output is
always an empty array, so it looks like it is me, but the invalid
variable name is an issue I think.

Regards,

Richard.

NOTE: The above is a simple test. I'm trying to map in nested data to
over 10 levels.
-- 
Richard Quadling.

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