Re: Re: Illegal string offset error in array

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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))  Â{
    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 ($subdir)) Â{
          if ($subfile != "." && $subfile != "..") {
          
            /*
            Â* This line does not work in PHP5 cause off an 
illegal string offset in an array ERROR
            Â*/
            $arrAvailableContent[$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))  Â{
    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 ($subdir)) Â{
          if ($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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