[SOLVED] Complete noob question - extract middle of filename

[richrock]

Right, today has been my first day of doing some coding using regex, and boy am I stumped

 

I've got a script to read a image folder, and need to extract parts of the filename in order to display information dynamically.  The images are set with a particular naming convention:

 

my real name_location-year-imagecode.jpg : I can get the name, year and image code, but for some reason I just don't know enough to get teh location.  It's got an underscore to the left and hyphen to the right, so how would I get it?

 

I was thinking along the lines of this

 

$regex = preg_match('/([^_.$/-])/', $filename);
		echo $regex;

Which obviously doesn't work, as this is rather hard, and I'm very tired   Id like it to have something like:

 

$regex = preg_match('/([_wildcardlocation-])/', $filename);

 

Thanks if you can help

[jackpf]

$location = reset(explode('-', end(explode('_', $filename))));

That might work

[jackpf]

Yeah, it does, I just checked it. Enjoy.

[richrock]

beat me to it   Sometimes I pause when googling, ask here then find the answers been there all along  :-\

 

Thanks.  I did do it slightly differently, but will try this too, just to learn how this stuff works.

[jackpf]

Could you post your method?

Just so people looking at this thread in the future will know how to do it.

[nrg_alpha]

Given the format:

my real name_location-year-imagecode.jpg

 

(read - assuming the underscore is the only character within this which separates the real name from the location, and the dash is the only characters which separates the location from the year and the imagecode)... one could use preg_split:

 

$str = 'my real name_location-year-imagecode.jpg';
$arr = preg_split('#[_-]#', $str);
echo "<pre>".print_r($arr, true);

 

Output:

Array
(
    [0] => my real name
    [1] => location
    [2] => year
    [3] => imagecode.jpg
)

[richrock]

even better - many thanks!  Really should learn more regex