Extracting components of a filename

Question

I have a string str ="/Users/user/Desktop/task/U6342_Account_20150112.txt"

I need return only U6342_Account_20150112 as an array ['U6342','Account','20150112']

I did

str.split('/')[-1].gsub('.txt','').split('_')

which gives me ['U6342','Account','20150112']

Is there any more convenient solution to do it?

Answer 1

The string looks like a filename, so you should use the library functions for handling file paths instead of rolling your own.

File.basename(str, '.txt').split('_')

Answer 2

I really like 200_success's answer, and that's what I'd use for this specific situation, but here's an alternative that's flexible for other situations, using a regexp:

str.match(/([^\/_]+)_([^_]+)_([^_]+).txt/).captures

Each piece you want is captured by a parentheses group, and we turn it into the array you want by calling captures on the match result.

• Nakilon: Flexible? I see only three captures. You probably want recursive regex.
• Jonah: good point, basically i was just saying regex + captures solves lots of these kinds of problems