Re: [PHP] Variable variables into an array.
On 10 August 2010 18:08, Andrew Ballard aball...@gmail.com wrote:
On Tue, Aug 10, 2010 at 12:23 PM, Richard Quadling rquadl...@gmail.com
wrote:
On 10 August 2010 16:49, Jim Lucas li...@cmsws.com wrote:
Richard Quadling wrote:
Hi.
Quick set of eyes needed to see what I've done wrong...
The following is a reduced example ...
?php
$Set = array();
$Entry = 'Set[1]';
$Value = 'Assigned';
$$Entry = $Value;
print_r($Set);
?
The output is an empty array.
Examining $GLOBALS, I end up with an entries ...
[Set] = Array
(
)
[Entry] = Set[1]
[Value] = Assigned
[Set[1]] = Assigned
According to http://docs.php.net/manual/en/language.variables.basics.php,
a variable named Set[1] is not a valid variable name. The [ and ] are
not part of the set of valid characters.
In testing all the working V4 and V5 releases I have, the output is
always an empty array, so it looks like it is me, but the invalid
variable name is an issue I think.
Regards,
Richard.
NOTE: The above is a simple test. I'm trying to map in nested data to
over 10 levels.
For something like this, a string that looks like a nested array reference,
you might need to involve eval for it to derive that nested array.
I'm happy with that.
It seems variable variables can produce variables that do not follow
the same naming limitations as normal variables.
It would seem so. If eval() works, can you rearrange the strings a
little to make use of parse_str() and avoid the use of eval()?
Andrew
php -r parse_str('a[1][2][3]=richard quadling'); var_dump($a);
outputs ...
array(1) {
[1]=
array(1) {
[2]=
array(1) {
[3]=
string(16) richard quadling
}
}
}
Perfect.
Thanks.
--
Richard Quadling.
--
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RE: [PHP] Variable variables into an array.
From: Richard Quadling Quick set of eyes needed to see what I've done wrong... The following is a reduced example ... ?php $Set = array(); $Entry = 'Set[1]'; ^^ Shouldn't that be $Set[1]? $Value = 'Assigned'; $$Entry = $Value; print_r($Set); ? Bob McConnell -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Variable variables into an array.
On 11 August 2010 13:58, Bob McConnell r...@cbord.com wrote: From: Richard Quadling Quick set of eyes needed to see what I've done wrong... The following is a reduced example ... ?php $Set = array(); $Entry = 'Set[1]'; ^^ Shouldn't that be $Set[1]? $Value = 'Assigned'; $$Entry = $Value; print_r($Set); ? Bob McConnell No. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Variable variables into an array.
Hi. Quick set of eyes needed to see what I've done wrong... The following is a reduced example ... ?php $Set = array(); $Entry = 'Set[1]'; $Value = 'Assigned'; $$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. -- Richard Quadling. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Variable variables into an array.
Richard Quadling wrote: Hi. Quick set of eyes needed to see what I've done wrong... The following is a reduced example ... ?php $Set = array(); $Entry = 'Set[1]'; $Value = 'Assigned'; $$Entry = $Value; print_r($Set); ? The output is an empty array. Examining $GLOBALS, I end up with an entries ... [Set] = Array ( ) [Entry] = Set[1] [Value] = Assigned [Set[1]] = Assigned According to http://docs.php.net/manual/en/language.variables.basics.php, a variable named Set[1] is not a valid variable name. The [ and ] are not part of the set of valid characters. In testing all the working V4 and V5 releases I have, the output is always an empty array, so it looks like it is me, but the invalid variable name is an issue I think. Regards, Richard. NOTE: The above is a simple test. I'm trying to map in nested data to over 10 levels. For something like this, a string that looks like a nested array reference, you might need to involve eval for it to derive that nested array. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Variable variables into an array.
On 10 August 2010 16:49, Jim Lucas li...@cmsws.com wrote: Richard Quadling wrote: Hi. Quick set of eyes needed to see what I've done wrong... The following is a reduced example ... ?php $Set = array(); $Entry = 'Set[1]'; $Value = 'Assigned'; $$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. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Variable variables into an array.
On Tue, Aug 10, 2010 at 12:23 PM, Richard Quadling rquadl...@gmail.com wrote: On 10 August 2010 16:49, Jim Lucas li...@cmsws.com wrote: Richard Quadling wrote: Hi. Quick set of eyes needed to see what I've done wrong... The following is a reduced example ... ?php $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 General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
