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]

Erik Andersson wrote:

<snip>

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

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

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

[pojo]

Uzytkownik "Zurab Davitiani" <ag*@mindless.com> napisal w wiadomosci
news:MZ*******************@newssvr25.news.prodigy. com...

Vamat certification wrote:

<snip>

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

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?

That syntax is not valid in PHP4. You may have a better luck with PHP5.

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

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

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

Shawn Wilson" <sh***@glassgiant.com> wrote in message
news:3F***************@glassgiant.com... Sticks wrote:
[re-ordered you post because top-posting is evil :o)]

<xyzzy> wrote in message news:Xf********************@comcast.com...

"Sticks" <sa**********@yahoo.com> wrote in message
news:3f********@usenet.per.paradox.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 the > resulting output to html files. this is necessary as the nature of the > hosting available to me for this particular page prohibits me from using > 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 include,

> because the script only works when it is in the same directory as the site.
> this arrangement does not suit me, as i would rather be able to store

the

> script in a separate folder.
>
>
>
>
>
My first thought is to create an array with your directory names that

you want to scan

$allDirs =

array("/usr/local/apache/htdocs","/usr/local/apache/site1/","/usr/local/apac

he/site1/htdocs") etc...

then iterate through the array and subsequent directory files.

while (list(,$dirName) = each($allDirs) {
// create an array of all files
$allFiles = scandir($dirName);
while (list ($key,$fileName) = each($allFiles) {
// each file within each directory
$page_buffer to $fileName stuf......
}
}

perhaps i havent been clear enough
i dont want to use includes to perform this task. i would like to get

the html output of a php script that is quite complicated and uses a variety of includes of its own which i do not wish to rewrite unless i absolutely have 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 for executing php scripts.

I'm not sure I understand your problem, but if you want to execute php

scripts on Linux, you can use

/pathinfo/php -q /pathinfo/scriptname.php

from the command line or a cron job. The "-q" suppresses HTTP headers. Or you 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 careful with this, though, as it's easy to overwrite the wrong files...