UNIX for Dummies Questions & Answers

-rw-r--r-- 1 fxpbftp fusion 368 Jun 10 08:34 FX_1.11840235236594E12.111234236809956

If I have a long list of files that look like this (they al begni with FX_1.#######.####) Sometimes, there may be less numbers or more in the filename, that varies.

I wish to isolate just the filename from this LS... how would that look like?

do I use

ls -al | cut -c#-# ? Is the LS always a fixed length (up to the filename)?

If you only want the filename why not just use ls without flags?

not withstanding DogDay's question.....

[assuming the files don't have embedded spaces in their names]

Code:

echo '-rw-r--r-- 1 fxpbftp fusion 368 Jun 10 08:34 FX_1.11840235236594E12.111234236809956' | sed 's/.* \([^ ][^ ]*\)$/\1/'

I could use LS but, I guess I should've said what I wanted to do.
Later on, I need to cut up the rest of the LS, such as the date of the file's submission and perform some let expr operations on them.

Thank you.

ls -l | awk '{print $9}'

if you want to go awk route....
[once again - assuming no embedded spaces in the file names]

ls -l | nawk '{print $NF}'

The default delimiter in AWK is [:space:] but the problem is.. (the post didn't paste all the spaces). Between some of the fields, there are 2 spaces, sometimes even 4 spaces.. etc..

Can awk identify variable space delimiters? This is why I didn't want to use awk for the LS. Correct or incorrect?

And yes, you're correct, there are no spaces in filenames.