Moving up a directory code

[newbiePHP]

Hey guys,

 

I have a engine that finds a file in a directory and opens it.

 

The file i want to open is in C:\temp. The PHP file is located in C:\wamp\www.

 

In the code when i set the directory to C:/temp/. The program runs fine.

 

When I set the directory to ../../temp/ it doesn't run.

 

In order to go up a directory is the command ../ and do I need to import something to make it work?

 

Thanks for any help!!

[jonsjava]

I tested it using XAMPP, and it worked just fine for me:

HTDOC ROOT:

c:\xampp\htdocs\jobsite2\test.php

test file:

c:\Temp\test.php

CONTENTS OF jobsite2/test.php:

<?php
$server_root = "../../../temp/";
include($server_root."test.php");

?>

CONTENTS OF temp/test.php:

<?php
print "it worked!";
?>

OUTPUT RECEIVED:

it worked!

So, my test worked fine. I don't see why you would have any issues doing the same.

[newbiePHP]

jonsjava,

 

thanks for the reply.

 

Would the fact that i'm running it through firefox be a problem?

[jonsjava]

It's server side. that won't matter.

[newbiePHP]

I tried compling it through the cmd and it gave this error

 

'..' is not recognized as an internal or external command, operable program or batch file.

 

Any ideas?

[jonsjava]

can you post your code, please?

[newbiePHP]

If i were to change the $directory to "C:/Temp" It works fine.

 

<?php

$FindFile = "Example";

$directory = '../../Temp/'; //trailing slash

$flag = false;
$ext = array( '.jpg' , '.gif' , '.txt' );
for( $i = 0; count( $ext ) > $i; $i++ )
{

     if( file_exists( $directory . $FindFile . $ext[$i] ) )
     {
          $flag = true;
          $name = $directory . $FindFile . $ext[$i];
          break;
     }
}

if( $flag == true )
{
	 exec($name);
}
?>

[jonsjava]

you're trying to run an image?

 

I did a slight mod of your script (changed example to test, had it go one more level up, and have it print the name of the file instead of executing it):

<?php
$FindFile = "test";
$directory = '../../../Temp/'; //trailing slash
$flag = false;
$ext = array( '.php', '.jpg' , '.gif' , '.txt' );
for( $i = 0; count( $ext ) > $i; $i++ )
{
     if( file_exists($directory.$FindFile.$ext[$i] ) )
     {
          $flag = true;
          $name = $directory . $FindFile . $ext[$i];
          break;
     }
}
if( $flag == true )
{
	 print $name;
}
?>

it worked, for what it is.  Only problem I see is you can't just

exec($name);

because going off the script you posted, you're trying to execute an image. Windows (as well as linux, and every other system in the world) can't execute an image.  You also can't execute a php file in windows, unless your system knows how to execute a .php extension.

[newbiePHP]

No example is a text file and the whole program runs fine and opens that text file when the directory is set to C:/temp

 

I just have those other extensions as options to play around with if i get .txt working. Thanks for the info about the images though.

 

I don't really see any difference between your script and mine so can't figure out why yours works and mine doesn't.

 

Do you know what this error means?

 

'..' is not recognized as an internal or external command, operable program or batch file

[jonsjava]

it means that it's trying to run the ".." as if it were a file, not a directory structure.

[newbiePHP]

Do you have some sort of special settings that allows your compiler to recongise it as a directory rather than a file?

 

Thanks for all the help by the way 

[wildteen88]

PHP does not compile scripts. How are you getting the following error:

'..' is not recognized as an internal or external command, operable program or batch file

This is not an error from PHP itself but a Windows command error. Which suggests to me that you are not actually passing the code to the PHP interpreter.

[jonsjava]

oops, I should have explained what I meant by "it". sorry for the confusion.