PHP Dir list

**Wynd** · Nov 1st, 2001, 08:40 PM

PHP Code:


<?

$dir_name = "C:\site";

$dir = opendir($dir_name); 

$file_list = "<ul>";



while ($file_name = readdir($dir))

{ 

  if (($file_name != ".") && ($file_name != ".."))

  { 

    $file_list .= "<li>$file_name</li>"; 

  }     

}



$file_list .= "</ul>";



closedir($dir); 

?>



<html>

<head>

<title>Files in <? echo "$dir_name"; ?></title>

Files in: <? echo "$dir_name"; ?>

<? echo "$file_list"; ?>

</body>

</html>

Is there a way I can just put this in a directory and have it list the files for that dir, and not have to manually specify the dir to list the files of?

denniswrenn · Nov 1st, 2001, 09:59 PM

try this:

$dir_name="./";

**da_silvy** · Nov 2nd, 2001, 02:03 AM

I don't think you would get the wanted effect for the $dir_name printout then....

But other than that, it works

**Wynd** · Nov 2nd, 2001, 09:03 PM

Yeah, well, the bad comes with the good but oh well. Thanks

**da_silvy** · Nov 2nd, 2001, 09:48 PM

What about having good with good:

This should be what you are looking for:

Did you want the absolute path or just directory name?

Absolute Path:

PHP Code:


<? 

$dir_name = "./"; 

$dir = opendir($dir_name); 

$file_list = "<ul>"; 



while ($file_name = readdir($dir)) 

{ 

if (($file_name != ".") && ($file_name != "..")) 

{ 

$file_list .= "<li>$file_name</li>"; 

} 

} 



$file_list .= "</ul>"; 



closedir($dir); 

?> 



<html> 

<head> 

<title>Files in <? echo getcwd(); ?></title> 

Files in: <? echo getcwd(); ?> 

<? echo "$file_list"; ?> 

</body> 

</html>

**da_silvy** · Nov 2nd, 2001, 09:53 PM

Just the directory's name:

PHP Code:




<? 

$dir_name = "./"; 

$dir = opendir($dir_name); 

$file_list = "<ul>"; 



while ($file_name = readdir($dir)) 

{ 

if (($file_name != ".") && ($file_name != "..")) 

{ 

$file_list .= "<li>$file_name</li>"; 

} 

} 



$file_list .= "</ul>"; 



closedir($dir); 

$dir = substr(strrchr(getcwd(), "\\"), 1);

?> 



<html> 

<head> 

<title>Files in <? echo $dir; ?></title> 

Files in: <? echo $dir; ?> 

<? echo "$file_list";?> 

</body> 

</html>

**sail3005** · Nov 4th, 2001, 10:28 AM

how would you get the full path of the files?

**da_silvy** · Nov 4th, 2001, 06:43 PM

I think you mean showing

Bullet: c:\apache\htdocs\index.php?

Here you go:

PHP Code:


<? 

$dir_name = "./"; 

$dir = opendir($dir_name); 

$file_list = "<ul>"; 



while ($file_name = readdir($dir)) 

{ 

if (($file_name != ".") && ($file_name != "..")) 

{ 

$filename = getcwd()."\". $file_name;

$file_list .= "<li>getcwd()\$file_name</li>"; 

} 

} 



$file_list .= "</ul>"; 



closedir($dir); 

?> 



<html> 

<head> 

<title>Files in <? echo getcwd(); ?></title> 

Files in: <? echo getcwd(); ?> 

<? echo "$file_list"; ?> 

</body> 

</html>

**sail3005** · Nov 4th, 2001, 07:08 PM

thanks, but, i get an error with it.

**Wynd** · Nov 4th, 2001, 08:16 PM

No, what I meant was the relative path i guess. Like if I put the file in C:\site\folder\dir\images\ it should say "Files in images/", if I put it in C:\site\db\ it should say "Files in db/", etc.

**da_silvy** · Nov 4th, 2001, 08:26 PM

Originally posted by Wynd
No, what I meant was the relative path i guess. Like if I put the file in C:\site\folder\dir\images\ it should say "Files in images/", if I put it in C:\site\db\ it should say "Files in db/", etc.

Have a look at my second post with code in it.

That does what you want ^^^^ here.

Sail, what does the error say?

**da_silvy** · Nov 4th, 2001, 08:31 PM

Yeah, sorry sail, I didn't test the code that i posted before, try this:

PHP Code:


<? 

$dir_name = "./"; 

$dir = opendir($dir_name); 

$file_list = "<ul>"; 



while ($file_name = readdir($dir)) 

{ 

if (($file_name != ".") && ($file_name != "..")) 

{ 

$cwd = getcwd();

$file_list .= "<li>$cwd\\$file_name</li>"; 

} 

} 



$file_list .= "</ul>"; 



closedir($dir); 

?> 



<html> 

<head> 

<title>Files in <? echo getcwd(); ?></title> 

Files in: <? echo getcwd(); ?> 

<? echo "$file_list"; ?> 

</body> 

</html>

**sail3005** · Nov 5th, 2001, 04:38 PM

cool, works now, thanks.

**Wynd** · Nov 5th, 2001, 07:27 PM

Well that's partly what I want. I just want the script to output the directory name, not the whole path. (C:\path\dir\images produces "Files in images", C:\otherpath\scripts produces "Files in scripts", etc)

**da_silvy** · Nov 6th, 2001, 12:04 AM

Have you tried this code yet:

I posted it before.

PHP Code:


<? 

$dir_name = "./"; 

$dir = opendir($dir_name); 

$file_list = "<ul>"; 



while ($file_name = readdir($dir)) 

{ 

if (($file_name != ".") && ($file_name != "..")) 

{ 

$file_list .= "<li>$file_name</li>"; 

} 

} 



$file_list .= "</ul>"; 



closedir($dir); 

$dir = substr(strrchr(getcwd(), "\"), 1);

?> 



<html> 

<head> 

<title>Files in <? echo $dir; ?></title> 

Files in: <? echo $dir; ?> 

<? echo "$file_list";?> 

</body> 

</html>

It does what you want...

**Wynd** · Nov 6th, 2001, 06:59 PM

Thanks, it works. One thing though, you forgot to escape the backslash, that threw me for a loop till i figured it out

**da_silvy** · Nov 7th, 2001, 02:30 AM

Woops, I didn't actually, but the php tags did that.

If you quote my message, you will see that i didn't forget to escape it

**sail3005** · Nov 22nd, 2001, 01:28 AM

is there anyway to tell it how to sort the files?

Like by, date modified, size, name, etc?

**da_silvy** · Nov 22nd, 2001, 01:53 AM

can't test it, but you should be able to load it into an array then sort the array...

**sail3005** · Nov 22nd, 2001, 09:29 AM

how would i load it into an array?

**sail3005** · Nov 23rd, 2001, 02:40 PM

any ideas please?

**da_silvy** · Nov 23rd, 2001, 07:16 PM

I think it's pretty difficult

I will have heaps of free time in a week, i'll figure something out, so just PM me.

after thursday - (no more exams

)

**sail3005** · Nov 23rd, 2001, 09:36 PM

Alright, i'll remember.

**da_silvy** · Nov 23rd, 2001, 10:09 PM

good for you

i think it would be quite difficult

can't remember if php can handle multiple dimension arrays

**sail3005** · Nov 23rd, 2001, 11:37 PM

that would be bad if it didn't. it would certainly make things difficult

**da_silvy** · Nov 24th, 2001, 01:25 AM

i was thinking though

It could be put into a normal array, depending on what order they want

i.e.

By FIlename

then the string =

"filename, size, " etc

By size

then the string =

"size, filename, " etc

then you sort the array
and split it

**sail3005** · Nov 24th, 2001, 11:49 AM

sounds good. i still have no idea how to do it though.

**da_silvy** · Nov 27th, 2001, 04:27 AM

here you go.

hope that works.

PHP Code:


<?

$dir_name = "./";

$dir = opendir($dir_name);

$myint = -1;

while ($file_name = readdir($dir))

{

if (($file_name != ".") && ($file_name != "..") && !is_dir($file_name))

{

$myint++;

$file_list[0][$myint] = "$file_name";

$file_list[1][$myint] = filesize($file_name);

$file_list[2][$myint] = filemtime($file_name);

}

}

//asort($file_list[0]); Sort by filename

//asort($file_list[1]); Sort by filesize

//asort($file_list[2]); Sort by modified time

asort($file_list[1]);

while (list($key) = each($file_list[1])) {

print

$file_list[0][$key].",".$file_list[1][$key].",".$file_list[2][$key]."<br>";

}

?>

**Wynd** · Nov 27th, 2001, 06:38 PM

The only part that works is the file size part. And, the filename and date parts print in the same order.

**Wynd** · Nov 27th, 2001, 06:47 PM

Nm, got it to work. I have been looking for this for a long time. Thanks for the help!

**da_silvy** · Nov 28th, 2001, 01:53 AM

No probs, what was the problem?

Did you quote the message and then copy that source or what?

**Wynd** · Nov 28th, 2001, 06:23 PM

I just c&p'ed ity from the web page. In this part:

PHP Code:


each($file_list[1])

you need to change "1" to 0, 1, or 2 depending on what you want.

**da_silvy** · Nov 29th, 2001, 03:01 AM

oh yeah, woops i should've mentioned that in my other post :|

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