OO - Accessing method in returned class without instance variable

[Erik Andersson]

Hi!

<?
class Foo {
function getMessage() {
return "foo";
}
}
class Bar {
function getFoo() {
return new Foo();
}
}
$bar = new Bar();
echo $bar->getFoo()->getMessage() ;
?>

Running this you will get a parse error on the line with:

echo $bar->getFoo()->getMessage() ;

In an object oriented manner I would like to argue that this expression is
completely valid. However it won't work in PHP and it annoys me. Is there
any other way of doing it without using a variable?
i.e.
$foo = $bar->getFoo();
$foo->getMessage() ;

The reason I don't want to use another variable is that I want to be able to
say:

<?=$bar->getFoo()->getMessage()?> .

All the best,

--
Erik Andersson ( E.N.E IT - Australia )

http://www.eneit.com/

- Australian, high quality web hosting from AU$4.95, all features included.

[Zurab Davitiani]

Re: OO - Accessing method in returned class without instance variable

Erik Andersson wrote:

<snip>[color=blue]
> echo $bar->getFoo()->getMessage() ;[/color]
<snip>[color=blue]
> Running this you will get a parse error on the line with:
>
> echo $bar->getFoo()->getMessage() ;
>
> In an object oriented manner I would like to argue that this expression is
> completely valid. However it won't work in PHP and it annoys me. Is there
> any other way of doing it without using a variable?[/color]

That syntax is not valid in PHP4. You may have a better luck with PHP5.
[color=blue]
> The reason I don't want to use another variable is that I want to be able
> to say:
>
> <?=$bar->getFoo()->getMessage()?> .[/color]

You have to use 2 statements instead of 1, like you guessed.

[pojo]

Re: OO - Accessing method in returned class without instance variable


Uzytkownik "Zurab Davitiani" <agt@mindless.c om> napisal w wiadomosci
news:MZVDb.7225 2$9J3.11965@new ssvr25.news.pro digy.com...[color=blue]
> Vamat certification wrote:
>
> <snip>[color=green]
> > echo $bar->getFoo()->getMessage() ;[/color]
> <snip>[color=green]
> > Running this you will get a parse error on the line with:
> >
> > echo $bar->getFoo()->getMessage() ;
> >
> > In an object oriented manner I would like to argue that this expression[/color][/color]
is[color=blue][color=green]
> > completely valid. However it won't work in PHP and it annoys me. Is[/color][/color]
there[color=blue][color=green]
> > any other way of doing it without using a variable?[/color]
>
> That syntax is not valid in PHP4. You may have a better luck with PHP5.
>[color=green]
> > The reason I don't want to use another variable is that I want to be[/color][/color]
able[color=blue][color=green]
> > to say:
> >
> > <?=$bar->getFoo()->getMessage()?> .[/color]
>
> You have to use 2 statements instead of 1, like you guessed.[/color]
Shawn Wilson" <shawn@glassgia nt.com> wrote in message
news:3FE07492.E C411FE6@glassgi ant.com...[color=blue]
> Sticks wrote:
> [re-ordered you post because top-posting is evil :o)][color=green]
> > <xyzzy> wrote in message news:XfqdnbU1EP quUEKi4p2dnA@co mcast.com...[color=darkred]
> > >
> > > "Sticks" <sarcasmic_98@y ahoo.com> wrote in message
> > > news:3fde9f74$1 @usenet.per.par adox.net.au...
> > > > ok... im not quite sure how to describe my problem.
> > > > i have a php script that runs through my entire php site and writes[/color][/color][/color]
the[color=blue][color=green][color=darkred]
> > > > resulting output to html files. this is necessary as the nature of[/color][/color][/color]
the[color=blue][color=green][color=darkred]
> > > > hosting available to me for this particular page prohibits me from[/color][/color][/color]
using[color=blue][color=green][color=darkred]
> > > > php/mysql as i would like.
> > > >
> > > > my script works simply by using output buffers and an include
> > > > the relevant section of code is as follows:
> > > >
> > > > ob_start();
> > > > include($source .$pagename);
> > > > $page_buffer = ob_get_contents ();
> > > > ob_end_clean();
> > > >
> > > > i was wondering if there is a way i can do this without using the[/color]
> > include,[color=darkred]
> > > > because the script only works when it is in the same directory as[/color][/color][/color]
the[color=blue][color=green][color=darkred]
> > > site.
> > > > this arrangement does not suit me, as i would rather be able to[/color][/color][/color]
store[color=blue][color=green]
> > the[color=darkred]
> > > > script in a separate folder.
> > > >
> > > >
> > > >
> > > >
> > > >
> > > My first thought is to create an array with your directory names that[/color][/color][/color]
you[color=blue][color=green][color=darkred]
> > > want to scan
> > >
> > > $allDirs =
> > >[/color]
> >[/color][/color]
array("/usr/local/apache/htdocs","/usr/local/apache/site1/","/usr/local/apac[color=blue][color=green][color=darkred]
> > > he/site1/htdocs") etc...
> > >
> > > then iterate through the array and subsequent directory files.
> > >
> > > while (list(,$dirName ) = each($allDirs) {
> > > // create an array of all files
> > > $allFiles = scandir($dirNam e);
> > > while (list ($key,$fileName ) = each($allFiles) {
> > > // each file within each directory
> > > $page_buffer to $fileName stuf......
> > > }
> > > }
> > >
> > >[/color]
> > perhaps i havent been clear enough
> > i dont want to use includes to perform this task. i would like to get[/color][/color]
the[color=blue][color=green]
> > html output of a php script that is quite complicated and uses a variety[/color][/color]
of[color=blue][color=green]
> > includes of its own which i do not wish to rewrite unless i absolutely[/color][/color]
have[color=blue][color=green]
> > to. these includes are disturbed when my script resides in a different
> > directory to the page.
> >
> > as it stands, the script works. it just isnt an ideal solution.
> > i am looking for something along the lines of the exec() command, only[/color][/color]
for[color=blue][color=green]
> > executing php scripts.
> >[/color]
>
> I'm not sure I understand your problem, but if you want to execute php[/color]
scripts[color=blue]
> on Linux, you can use
>
> /pathinfo/php -q /pathinfo/scriptname.php
>
> from the command line or a cron job. The "-q" suppresses HTTP headers.[/color]
Or you[color=blue]
> could do:
>
> /pathinfo/php -q /pathinfo/scriptname.php > index.html
>
> to send all output from that php script to a file called index.html. Be[/color]
careful[color=blue]
> with this, though, as it's easy to overwrite the wrong files...[/color]