Bit more complicated PHP question

[JmZ]

Do you mean one gz contains many files? or may gzips contain one file each?

If they each have one file, you just loop through them or stick them in an object.

for example, off the top of my head:

$files = array('some.gz', 'file.gz', 'made.gz', 'up.gz');

function hostImage($filename) {
    $file = gzfile($filename);
    $file = implode("\n",$file);

    $matches = array(
        array('rapidshare.png', '[url=http://rapidshare'),
        array('megaupload.png', '[url=http://megaupload')
    );

    $image = 'none.png';

    foreach($matches as $match) {
        if(strpos($file, $match[1]) !== false) {
            $image = $match[0];
            break;
        }
    }

    return $image;
}

foreach($files as $file) {
     $image = hostImage($file);
     // do other things here
} 

It'd be nicer class based now I believe, because having the array of strings to match inside the function isn't ideal.

Also, this will return 'none.png' if nothing is found.

[Porsche_maniak]

@JmZ

Yea .. They are arround 1300 .gz files in a folder and keep increasing . Each .gz contains only 1 .txt file . I have a path where the .gz files are - $pathz = CONTENT_DIR.$y.'/'.$m;
The path is taking 10 .gz files.When user click the next page it takes the next 10 .gz.

I hope i haven't confused you.

[JmZ]

Yeah I see.

Well you'd want to find all gz files in the dir first, then loop through like above.

e.g.

change $files at the top of the code, to:

$current_dir = getcwd();
chdir($pathz);
$files = glob('*.gz');
chdir($current_dir); 

[CyberHacK]

Coding Geek battle =/

[Porsche_maniak]

It is content/10/05
I tried

$pathz = CONTENT_DIR.$y.'/'.$m.'/'.$glo;

 $glo=glob('*.gz');

 
 echo $pathz;

To see if i am going to take the .gz file names but it was again content/10/05/

[litewarez]

/*
    *Functions
*/
function hostImage($filename) {
    $file = gzfile($filename);
    $file = implode("\n",$file);

    $matches = array(
        array('rapidshare.png', '[url=http://rapidshare'),
        array('megaupload.png', '[url=http://megaupload')
    );

    $image = 'none.png';

    foreach($matches as $match) {
        if(strpos($file, $match[1]) !== false) {
            $image = $match[0];
            break;
        }
    }
    return $image;
}

//Ditrectory stuff
$gdir = CONTENT_DIR.$y.'/'.$m.'/*.gz';

foreach(glob($gdir) as $file)
{
    $image = hostImage(CONTENT_DIR.$y.'/'.$m.'/' . $file);
    echo $image; // rapidshare.png or megaupload.png
} 

[Porsche_maniak]

Warning: gzfile(content/10/05/content/10/05/entry100501-210451.txt.gz) [function.gzfile]: failed to open stream: No such file or directory

the bolded shouldnt exist..

[Lock Down]

Wow what a blob of too many coders.

First let me say I have no idea where your code came from as I don't see content_dir being defined anywhere.
But looking at what it is producing try removing the content_dir. from the command whatever it was.
Seeing what your total code is for these routines would be nice since you are following 2 peoples advice.

[litewarez]

no:

Chnage

$image = hostImage(CONTENT_DIR.$y.'/'.$m.'/' . $file);

to

$image = hostImage($file);

i forgot glob returns an absolute path

[Porsche_maniak]

Thanks guys !
I managed to get it working...