Re: [PHP] Troubles from the newb again
Jeff Taylor wrote:
Yeah I have, but what I dont understand is that the value of $name is an
object, and was accepted in the original array as the index:
the original array was:
$me = new Toon(,xxx,xxx,,xxx,etc,etc,etc)
$him = new Toon(xxx,xx,x,etc,etc,etc,etc)
$array=array($me,$him)
Hell no,
there's a difference between:
$array[$me] and array($me);
The first results in:
$array[$me] = null;
the other results in
$array[0] = $me;
Note that in the 2nd one you assign $me as the first value of the array,
and in the 1st one you assign $me to be the first index/key of the array
(with a null value).
So this new array, I want to isolate those toons still alive:
if alive then
add toon to new array
Is this possible in anyway?
The reason why I dont want it to put it in a value:
foreach ($newarray as $name)
{
if (in_array($name,$characters))
{
$newarray[$name] = 1;
}
else
{
$newarray[$name] = 2;
}
}
Sorry for the spam everyone
Then use $name as being an actual NAME (aka "string") instead of the
entire `toon` you are talking about (aka "object").
- tul
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Re: [PHP] Troubles from the newb again
On Sat, 2007-03-10 at 16:39 +1100, Ryan Fielding wrote: > Robert Cummings wrote: > > On Sat, 2007-03-10 at 15:31 +1030, Jeff Taylor wrote: > > > >> Hey everyone, > >> Newb back again - Im trying to populate my arrays, but getting this error > >> again: > >> Warning: Illegal offset type in > >> > > > > This usually means that you are using a non-scalar value as an index. > > Use var_dump( $array ) to see what's in there. You will probably find > > one of the following: an array, and object, or the null value. > > > > Cheers, > > Rob. > > > A good habit to get into is always using if(is_array()) when using foreach. Sure, if you don't know what is in the array and are quite happy to waste cycles calling a function on every single value :( ... In other words... HELL NO! You shouldn't have to filter your data if it should already be in an expected format. If it isn't in the expected format then you have a bug -- fix the bug instead of "handling" it within every foreach loop that touches the data. Cheers, Rob. -- .. | InterJinn Application Framework - http://www.interjinn.com | :: | An application and templating framework for PHP. Boasting | | a powerful, scalable system for accessing system services | | such as forms, properties, sessions, and caches. InterJinn | | also provides an extremely flexible architecture for | | creating re-usable components quickly and easily. | `' -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Troubles from the newb again
Yeah I have, but what I dont understand is that the value of $name is an
object, and was accepted in the original array as the index:
the original array was:
$me = new Toon(,xxx,xxx,,xxx,etc,etc,etc)
$him = new Toon(xxx,xx,x,etc,etc,etc,etc)
$array=array($me,$him)
So this new array, I want to isolate those toons still alive:
if alive then
add toon to new array
Is this possible in anyway?
The reason why I dont want it to put it in a value:
foreach ($newarray as $name)
{
if (in_array($name,$characters))
{
$newarray[$name] = 1;
}
else
{
$newarray[$name] = 2;
}
}
Sorry for the spam everyone
"Robert Cummings" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> On Sat, 2007-03-10 at 15:31 +1030, Jeff Taylor wrote:
> > Hey everyone,
> > Newb back again - Im trying to populate my arrays, but getting this
error
> > again:
> > Warning: Illegal offset type in
>
> This usually means that you are using a non-scalar value as an index.
> Use var_dump( $array ) to see what's in there. You will probably find
> one of the following: an array, and object, or the null value.
>
> Cheers,
> Rob.
> --
> ..
> | InterJinn Application Framework - http://www.interjinn.com |
> ::
> | An application and templating framework for PHP. Boasting |
> | a powerful, scalable system for accessing system services |
> | such as forms, properties, sessions, and caches. InterJinn |
> | also provides an extremely flexible architecture for |
> | creating re-usable components quickly and easily. |
> `'
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Re: [PHP] Troubles from the newb again
Robert Cummings wrote: On Sat, 2007-03-10 at 15:31 +1030, Jeff Taylor wrote: Hey everyone, Newb back again - Im trying to populate my arrays, but getting this error again: Warning: Illegal offset type in This usually means that you are using a non-scalar value as an index. Use var_dump( $array ) to see what's in there. You will probably find one of the following: an array, and object, or the null value. Cheers, Rob. A good habit to get into is always using if(is_array()) when using foreach. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Troubles from the newb again
On Sat, 2007-03-10 at 15:31 +1030, Jeff Taylor wrote: > Hey everyone, > Newb back again - Im trying to populate my arrays, but getting this error > again: > Warning: Illegal offset type in This usually means that you are using a non-scalar value as an index. Use var_dump( $array ) to see what's in there. You will probably find one of the following: an array, and object, or the null value. Cheers, Rob. -- .. | InterJinn Application Framework - http://www.interjinn.com | :: | An application and templating framework for PHP. Boasting | | a powerful, scalable system for accessing system services | | such as forms, properties, sessions, and caches. InterJinn | | also provides an extremely flexible architecture for | | creating re-usable components quickly and easily. | `' -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
