help with this code please

[WebStyles]

and don't you thing that learning the very basics is a good idea before starting to code a project ?

 

[wkdw1ll1ams]

all i need to is add some php code so i dont have to hyperlink the film directories and be here all day  so, the avi directory is

 

films/movie file/movie.avi

[WebStyles]

yeah man, I know what you need, but the purpose of this forum is to help programmers understand and solve problems. You don't seem to know any programming, and you don't even want to understand it, you basically just want someone to write the code for you (which has already been done, and still you couldn't get it to work)... so we're in a difficult spot right now... I don't mind explaining everything 20 times if necessary until you understand it, but you need to WANT to learn this stuff.

[wkdw1ll1ams]

i want to learn on the side aswell.

 

so, this checks the directory?

 

if(is_dir($dir.'/films'.$file)) {

 

$dir holds /films

what does $file mean?

[WebStyles]

if(is_dir($dir.'/films'.$file)) {

 

Better, but still wrong.

 

you need: directory/folder/filename.ext

 

each variable ($dir and $file) get replaced by their content, so in your code above, if $file holds the filename (plus extension), what would be the result of what you typed ($dir.'/films'.$file) ?

[wkdw1ll1ams]

you tell me? idk.

[WebStyles]

no man, believe me, you NEED to understand this... ok, same example as I gave before... when I do this:

 

$var = "whatever";
echo $var . ' xxx ' . $var . ' xxx';

 

if I tell you that echo just prints whatever comes after it on the screen, and $var gets replaced with the value that was assigned to it, what would be the result?

[wkdw1ll1ams]

ok, i understand echo because i am familiar with batch.

 

echo $var . ' DIR' . $var . ' filename.ext';

 

?

[WebStyles]

no. forget about your code for now, and just look at this:

 

$var = "whatever";
echo $var . ' xxx ' . $var . ' xxx';

 

what would be the result printed on the screen?

[wkdw1ll1ams]

whatever xxx whatever xxx

 

?

[WebStyles]

YES!

 

ok, what about this? (still ignore YOUR code, just look at the example)

 

$mainDirectory= './films';
$folder = 'Film1';
$filename = 'GoneWithTheWind.avi';
echo $mainDirectory . $folder . $filename;

[wkdw1ll1ams]

films/film1/GoneWithTheWind.avi

 

or

 

./filmsFilm1GoneWithTheWind.avi

 

i dont know if theres mean to be "/" in there or "." ?

[WebStyles]

wrong. please try again.

[wkdw1ll1ams]

./filmsFilm1GoneWithTheWind.avi

 

 

[WebStyles]

YES!

 

so how would you fix this line of code:

echo $mainDirectory . $folder . $filename;

so that the result is: ./filmsFilm1GoneWithTheWind.avi

 

??

[wkdw1ll1ams]

there is nothing to fix to make it ./filmsFilm1GoneWithTheWind.avi

 

 

your reply

so that the result is: ./filmsFilm1GoneWithTheWind.avi

 

my reply

./filmsFilm1GoneWithTheWind.avi

 

[WebStyles]

ok, cool. (it was a trick question)

 

now this:

echo $mainDirectory . '/' .$folder . '/' .$filename;

 

??

[wkdw1ll1ams]

im not sure you can have 2 echos? and there is nothing telling the php what $dir is?

[WebStyles]

sorry, was a typo. removed one of them.

 

and $dir doesn't exist in my code. (ignore your original code for goodness sake)

[wkdw1ll1ams]

oops i meant $maindirectory

[WebStyles]

[assume the same values as defined before]

[wkdw1ll1ams]

./films/Film1/GoneWithTheWind.avi

[WebStyles]

ok, so now that's beginning to look like the path you need right?..

 

so what would this output?

$dir = './films';
$file = 'SomeFile.avi';
echo $dir.'/films'.$file;

[wkdw1ll1ams]

./films/filmsSomeFile.avi

[WebStyles]

a-ha!

 

So you see how it is wrong in your code?