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How do I get only the file name from a listing?

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Top Forums Shell Programming and Scripting How do I get only the file name from a listing?
# 1  
Old 12-05-2009
How do I get only the file name from a listing?
Hi,
I am trying to get a file name only. Could anyone help me on the same.
Meaning I have a file say list.out which holds below output
./xyz/abc.txt
./xyz/hij.txt
I want an output as below
abc.txt
hij.txt
i.e I want to delete everything before abc.txt
Please help me out if any one has the ans.
# 2  
Old 12-05-2009
Please use a more useful title next time. This cannot possibly be that urgent - i.e., your system will not crash if you don't know how to do this.

The command you want is basename
Code:
>echo basename /path/to/file/file.txt
file.txt
>

# 3  
Old 12-05-2009
Code:
xargs -n1 < list.out basename
abc.txt
hij.txt

# 4  
Old 12-05-2009
cat list.out | awk -F "/" '{print $3}'
# 5  
Old 12-05-2009
Awk is incredibly much faster (more noticeable for larger list.out files) - and handles filenames with spaces, but I'd change it slightly

Code:
awk -F/ '{print $NF}' list.out

0.05 seconds for 8000+ files, versus 6.3 seconds using xargs.

(not that I'm a fan of the UUOC club - it took the same time with or without cat!)

It's even quicker than sed (0.09 seconds)
Code:
sed "s+.*/++" list.out

Last edited by Scott; 12-05-2009 at 01:08 PM..
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