[Bonzo]

I have a bit of code I am using to list all the files in a directory. But all I realy want are the "jpg" can anyone tell me either how to select the jpg only or how to delete everything apart from jpg from the output list.

Thanks Anthony

PHP Code:

function get_dirlist($start_dir) {

exec("ls -R $start_dir",$f_list);

$dir_str = $start_dir;

$filelist[0] = $start_dir; $i = 1;

for ($count=0; $count<count($f_list); $count++) {

   if ($f_list[$count] == "") { continue; }

   if (substr($f_list[$count],strlen($f_list[$count])-1,1) == ":") {

     $dir_str = substr($f_list[$count],0,strlen($f_list[$count])-1);

     $filelist[$i] = $dir_str;

     $i++;

   } else {

     $file_str = "$dir_str/$f_list[$count]";

     if (is_file($file_str)) {

       $filelist[$i] = $file_str;

       $i++;

     }

   }

}

return $filelist;

} 

[Stefan]

what about:

PHP Code:

function get_dirlist($start_dir) {

exec("ls -R $start_dir *.jpg",$f_list);

$dir_str = $start_dir;

$filelist[0] = $start_dir; $i = 1;

for ($count=0; $count<count($f_list); $count++) {

   if ($f_list[$count] == "") { continue; }

   if (substr($f_list[$count],strlen($f_list[$count])-1,1) == ":") {

     $dir_str = substr($f_list[$count],0,strlen($f_list[$count])-1);

     $filelist[$i] = $dir_str;

     $i++;

   } else {

     $file_str = "$dir_str/$f_list[$count]";

     if (is_file($file_str)) {

       $filelist[$i] = $file_str;

       $i++;

     }

   }

}

return $filelist;

} 

[Bonzo]

Thanks for the info Stefan but it had no effect.

I tried putting the results into an array and using ' *.jpg ' but was not sure that '*.jpg' could be used. Now I see it can I will have another go tonight.

Any other help would be gratfuly recived.

Anthony

[Bonzo]

I am still trying to get this to work and have found another method that I still cant get to work.

PHP Code:

$myDirectory = opendir("$image_path");



    while($entryName = readdir($myDirectory))

    {

    $dirArray[]=$entryName;

    }



 function select_png($select)

{

    return($select == '*.png');

}



// line to view what results I am getting ; currently Array ( )

 print_r(array_filter($dirArray, "select_png")); 

Can someone help me out ?

Anthony

[blaw]

Hi there,

Maybe you want to this.

PHP Code:

$myDirectory = opendir("$image_path");

while($entryName = readdir($myDirectory)) {

    if (substr($entryName, -3) == 'jpg') {

        $dirArray[]=$entryName;

    }

} 

Trick here is that you check if each file you're getting from the directory resource is a jpg file or not by looking at the last three characters of it (with substr()). It doesn't check the actual file type, so even if it's an myapp.exe file renamed as myapp.jpg, this code will pick it up.

Stefan's is to filter the non-jpg files BEFORE putting them into the directory resource, while mine is AFTER. Ideally, you should go with Stefan's, because it's less memory consuming, depending on how many non-jpg files you are looking at. (but mine also works on Windows...!)

[Bonzo]

Dont let NeverMind see this he will probably laugh himself to death !!!

This is a right old " mish mash " and there must be a betterway; I could not get your method to work blaw but I have come up with the below using it.

Anthony

PHP Code:

$myDirectory = opendir("$image_path");

// What I want is something to only read jpg here from the directory

while($entryName = readdir($myDirectory))

{

$dirArray[]=$entryName;

}



$indexcount = count($dirArray);



for ($i=2; $i<$indexcount; $i++ )

  {

    if ( substr($dirArray[$i], -3) == 'jpg')

    {echo $dirArray[$i]; } // Just to test the output

// Output is banner.jpgkatie.jpg.......etc.

// So I need to build another array from this - very longwinded way of doing it

  } 

[Bonzo]

Dont know what I was doing wrong but :

PHP Code:

$myDirectory = opendir("$image_path");

while($entryName = readdir($myDirectory)) {

    if (substr($entryName, -3) == 'jpg') {

        $dirArray[]=$entryName;

    }

}



closedir($myDirectory);



$indexcount = count($dirArray);

for ($i=0; $i<$indexcount; $i++ )

{ echo $dirArray[$i];}

// Output banner.jpgkatie.jpg......etc. 

Now works Anthony