Posted by Johannes Findeisen on 03/28/05 18:24

Hello Jasper and Rasmus,

thanks for reply! I have found out why this does not work. Take a look below.

On Monday 28 March 2005 02:33, Rasmus Lerdorf wrote:
> Jasper Bryant-Greene wrote:
> > Rasmus Lerdorf wrote:
> >>>> Jasper Bryant-Greene wrote:
> >>>>> You can't access string offsets with square brackets [] in PHP5.
> >>>>> You need to use curly braces {} instead.

I have replaced all square brackets with curly braces when using string
offsets. This didn't helped but worked on PHP4 and 5.

> > Yes, but he's talking about PHP5. He probably has E_STRICT on, which is
> > why he's getting that error.
> >
> > I was telling him to use the correct syntax, which will cause him to not
> > get that error.

I have E_STRICT on but the same behaviour with E_STRICT off

> I know we are talking about PHP5, and that syntax does not throw an
> E_STRICT error.
>

Right!!

Okay, lets see my original function again (copied from first post):

function getNavigationContent($content_path) {
    $i = 0;
    $dir = opendir($content_path);
    while ($file = readdir ($dir))    {
        if ($file != "." && $file != ".." && !in_array($file, explode(',',
$this->arrConf['default_filtered_files']))) {
            $arrAvailableContent[$i] = $file;
            if(is_dir($content_path . $file)) {
                echo $n = 0;
                $subdir = opendir($content_path . $file);
                while ($subfile = readdir ($subdir))  {
                    if ($subfile != "." && $subfile != "..") {
                   
                        /*
                         * This line does not work in PHP5 cause off an
illegal string offset in an array ERROR
                         */
                        $arrAvailableContent[$i]['submenu'][$n]['file'] =
$subfile;
                    $n++;
                    }
                }
                closedir($subdir);
            }
        $i++;
        }
    }
    closedir($dir);
    return $arrAvailableContent;
}

In the code take a look at the following line:

$arrAvailableContent[$i] = $file;

I am assiging a string to the var $arrAvailableContent[$i] .

Now lets take a look at the line i thought which was wrong:

$arrAvailableContent[$i]['submenu'][$n]['file'] = $subfile;

Now i am assigning a string to $arrAvailableContent[$i]['submenu'][$n]['file']
and this is what doesn't work. Now i have var_dump'ed all the arrays and i
found out, that there are no values in $arrAvailableContent[$i]['submenu']
[$n]['file'] in PHP4. In PHP5 the script exits because of a Fatal Error. But
if i set the line:

$arrAvailableContent[$i] = $file;

To:

$arrAvailableContent[$i]['file'] = $file;

Everything works fine in PHP4 and PHP5.

My function now looks like this:

function getNavigationContent($content_path) {
    $i = 0;
    $dir = opendir($content_path);
    while ($file = readdir ($dir))    {
        if ($file != "." && $file != ".." && !in_array($file, explode(',',
$this->arrConf['default_filtered_files']))) {
            $arrAvailableContent[$i]{'file'} = $file;
            if(is_dir($content_path . $file)) {
                echo $n = 0;
                $subdir = opendir($content_path . $file);
                while ($subfile = readdir ($subdir))  {
                    if ($subfile != "." && $subfile != "..") {
                   
                        $arrAvailableContent[$i]{'submenu'}[$n]{'file'} =
$subfile;
                    $n++;
                    }
                }
                closedir($subdir);
            }
        $i++;
        }
    }
    closedir($dir);
    return $arrAvailableContent;
}

This works in PHP4 (4.3.10) and PHP5 (5.0.3)!

Okay, after this my array doesn't look like the application wants it but this
is because i never had this problem before and it should be fixed at all. I
can change it quickly in my app since it is designed very clean but i just
wanted you to know my conclusion.

I have developed many lines of code in PHP4 in the last years but this weekend
i gave PHP5 a try. I must say i am impressed. PHP4 really is missing some
features but it is great too. Thanks Rasmus for inventing this scripting
language. I will become a better PHP programmer just only with the new
features in PHP5!

Have a nice day!

Regards
--
# Johannes Findeisen

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