finding out last version with bash

[aliov]

Hi all , i have the following situation, a directory contains something like :

linux v1r0p2
linux v3r7p2
linux v4r6p9

and so on, i want to write a script can grab me the last version, i've done something, but it a bit complicated and long, i'm wondering if there is a combination of sed and awd can do it easily.

[hugortega]

ls -1 | sort -r | head -1

(only if the version strings have the same length... I think)

[aliov]

ls -l  | grep -w 'linux' | sort -r | head -1

I use it like this and it seems it's working , thanks a lot .

Cheers.

[matonb]

If the latest version is always the last one created, add "t" to the ls command to sort by time as well, the "t" and "r" flags to ls imply sorting so you don't need to pipe it through the sort command:

ls -ltr linux* | head -1

Edit:
You don't need grep either...

[aliov]

If the latest version is always the last one created

This is not true, at least in my case, some times there is some modification done in older verions, that's why i didn't use -ltr flags.

[matonb]

In that case drop the "t"...

[scm]

If the latest version is always the last one created, add "t" to the ls command to sort by time as well, the "t" and "r" flags to ls imply sorting so you don't need to pipe it through the sort command:

ls -ltr linux* | head -1

To be exactly equivalent, you'd need

ls -lr *linux* | head -1

since grep will match 'linux' anywhere in the name, not just at the beginning. And, of course, you don't need the -l if you only want the filename - remember that ls writes filenames one per line unless it's writing to a terminal (so you don't need the -1 (that's a one, not an L) that hugortega used in his example).

Also be aware that it'll break after v9 since v10 will sort after v1 but before v2.