ltrim in while construct

andrewdunstall

Hi All,
I am using this code:

while ($element = each($name))
{
ETC
ETC

I need $name to be left trimmed.
HOW DO I nest ltrim inside the each function inside the while brackets?
ltrim has to execute on each iteration - but the following code snippet doesn't work:

while ($element = each(ltrim($name, $dir)))
{

Thanks in advance - Andrew

laserlight

Use a foreach loop.

foreach ($name as $key => $value)
{
    $name[$key] = ltrim($value);
}

andrewdunstall

Thanks laserlight,
But where do I define the bit to trim off?

I was using:
$name=ltrim($name, $dir);

$dir is the stuff I want to get rid of. Your example looks like this:
$name[$key] = ltrim($value);

I am using "glob" earlier in my code to give me some file names, but I'm also getting the name of the folder the files are in as the returned name. I don't want the name of the folder, I want to trim that off. I have googled around, but I'm stumped.
Could you elaborate on your example?

Andrew

NogDog

Based on your variable names, I'm wondering if perhaps the basename() function is what you need?

andrewdunstall

Thanks NogDog - Basename eh? I hadn't heard of that one. I also found str_replace to be an alternative.
I have changed my loop structure and I'm using ltrim.

So, the problem is solved.
Now on to the next looming code stopper.

Andrew

andrewdunstall

Opps, I spoke too soon.
If I use str_replace, it is changing the name of my files - which is bad.
The files I need to access are in a directory 4 deep, e.g.
clients/name/01/thumbnails

basename() is only returning 'thumbnails',
str_replace will delete the letters 'cl' off any file name that starts with 'cl', and
ltrim is also deleting 'cl' from the filenames ( I assume because the path starts with the name 'clients').

I'm sure there is a simple way for me to get:
photo01.jpg
when the full path is:
clients/name/01/thumbnails/photo01.jpg

All the examples I have seen say to use str_replace. But this now seems dangerous in light of what I have found out (see above).

Any other ideas?
Thanks - Andrew

NogDog

I'm probably missing something in the problem description, but basename() works just fine in this context:

$path = 'clients/name/01/thumbnails/photo01.jpg';
$file = basename($path);
echo $file;  // returns "photo01.jpg"