hello all,
iam new to shell scripting.I have searched the forum and could'nd find a close enough answer and hence this post:

I want to list all the file names whose names don't start as abc.

For example if my folder constains files with names: abc123.txt,erdf23.rdf,ed45r.fmb

i want a command which will list only files erdf23.rdf,ed45r.fmb and omit the file abc123.txt.

the ls command does'nt seem to have a -not option

If you have files in underlying directories and you want to omit them:

Perfect Franklin...Thank You very much

Sorry Frank ...If i have to elimiate more than one name in the file how do i do it?

i mean don't select files whose name starting with abc or efg

ls -ld !(abc*,efg*)|awk '!/^d/{print $NF}' don't seem to work

Try this:

Thanks Frank when I assign this code to a variable and try to print the variable contents it just prints nothing.

I do this:

FILELIST=`ls -l |awk '/^-/ && $NF !~ /^abc/ && $NF !~ /^efg/ && $NF ~ /.out/ {print $NF}'`
echo $FILELIST


Neither does this work:
FILELIST=`echo ls -l |awk '/^-/ && $NF !~ /^abc/ && $NF !~ /^efg/ && $NF ~ /.out/ {print $NF}'`
echo $FILELIST

Please help..Ideally it shold be printing the file name whose extension is .out and not begining with abc or efg

It should work, post the output of ls -l within code brackets.

Regards