[SOLVED] [bash] globbing

hashbang#! · 05-12-2011, 05:02 PM

Code:

cat ${PREFIX}[0-9][0-9][0-9][0-9] ${PREFIX}[0-9][0-9][0-9][0-9][0-9][0-9] >list

I am specifically looking for file names of ending in 4 or 6 digits. Can this be expressed in a smarter way?

sycamorex · 05-12-2011, 05:19 PM

This should work:

Code:

$ prefix="abcd"
$ ls | sed -nr "/$prefix([0-9]{4}$|[0-9]{6}$)/p"
abcd4444
abcd666666

hashbang#! · 05-12-2011, 05:30 PM

Brilliant! Just need one small correction or we're also catching files where $prefix is preceded by other characters:

Code:

ls | sed -nr "/^$prefix([0-9]{4}$|[0-9]{6}$)/p"

sycamorex · 05-12-2011, 05:33 PM

I'm glad that's what you wanted. Please don't forget to mark the thread solved.

hashbang#! · 05-12-2011, 06:09 PM

Oops - just noticed it's not happy about $PREFIX containing a relative pathname:

PREFIX=dir/fileprefix

sycamorex · 05-12-2011, 06:23 PM

Either change the prefix to escape the forward slash:

Code:

PREFIX=dir\/fileprefix

or change the delimiter to something else, eg:

Code:

ls | sed -nr ":^$prefix([0-9]{4}$|[0-9]{6}$):p"

hashbang#! · 05-13-2011, 03:23 AM

Now this is ingenious! We just need to tell sed to change the delimiter to get the above to work:

When using sed for substitution, the character following the 's' is automatically taken as the delimiter. For patterns the non-default delimiter needs to be introduced with a backslash:

Code:

ls | sed -nr "\:^$prefix([0-9]{4}$|[0-9]{6}$):p"

Thank you for your time, sycamorex.

sycamorex · 05-13-2011, 03:36 AM

Quote:

For patterns the non-default delimiter needs to be introduced with a backslash:

Indeed. I forgot about it.

hashbang#! · 05-13-2011, 05:00 AM

I spent 10 minutes reasearching sed options; I learnt more from that than I would have done had it worked out of the box!