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Grep filename in shell script

Posted by hycheah (hycheah), 23 July 2007
I have a folder that consists of the files below:

comment.txt
checkin.sh
filename1.list
filename2.list
list1.rpt
listb.rpt
hello3.pl

I need to create a new file, files.txt which is lists all the filenames except comment.txt and checkin.sh.

I tried
(ls | grep [^comment.txt|checkin.sh]) > files.txt
but it doesn't seem to work.

Please help. Thanks!

Posted by admin (Graham Ellis), 23 July 2007
The [ to ] element of a regular expression calls up any one character within the list, which is not what you want;   you want simply | to mean "or" to whle strings, and you should use egrep as not every basic grep does that in practise.   The -v option is used to give everything except matching lines.

That gives you a rewritten statement:
ls | egrep -v 'comment.txt|checkin.sh' > files.txt

Not perfect - the "." actually matches any character at all, and you are looking within the line reported by ls in each case and not at the whole line, but it should do for you.

Posted by george_Ball (george), 27 July 2007
You can just escape the '.' in the regular expression and then it's more or less complete...

ls | egrep -v 'comment\.txt|checkin\.sh' > files.txt




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