remove file

[Cheung_2]

I world like to remove some files but omit one or part of these files, how to do it?

eg.
$ ls
a1 a2 a3 a4 a5
a6 ab a8 a9 a10
a1b a2b

if I run ??? rm a* ??? , then all files will be removed , how can I omit to remove a2 & a6 and the file name ended with ???b???? Thx.

[Robert Thorneycroft]

The easiest way to do this is just tu use rm in interactive mode ie
rm -i a*
Then just answer y to the files you wish to remove.
This way will be 100 times quicker than trying to find some kind of pattern match.

Kind regards,

Robert Thorneycroft

[Michael Tully]

One of the easiest ways is to use:

# rm -i

[Ralph Grothe · Accepted solution]

Just a general hint before you do any deletions,
use the shell built-in echo command to see what the shell expands your wildcards (i.e. meta characters) to.

Maybe this is not quite what you want, but remember that you can exclude certain character sets with the '!'

e.g.

echo a[!26]*

However in your case a more subtle parsing is required, where I would use commands that can handle regular expressions (e.g. grep, sed, awk, perl)

[Leif Halvarsson_2]

Hi,
One method you can use when you want to do something with a large number of files (not necessary removing them) is to list to a file and use a text editor to process the file. Then you can use this file as input to a script. Example:

ls a* >tmpfile

vi tmpfile

while read file
do
#do something
done

[Niraj Kumar Verma]

#/bin/sh

for index in `ls a* |grep -v "a[26]"`
do
echo "== removing $index =="
rm -rf $index
done


-Niraj

[justin berkman]

hi guys



command :
rm a1 a[3-5] a[8-10]

omit to remove a2 - a6 - a1b and a2b

[Jean-Louis Phelix]

hi,

egrep (or grep -E) with multiple patterns could do it.

rm -i $(ls a* | egrep -v 'a2|a6|b$')

regards

[Robert Thorneycroft]

Jimmy your substitution [8-10] will not work as the value in brackets will only substitute a single character, which 10 is not.

[justin berkman]

hi guys

the correct command is :

rm a1 a[3-5] a[8-9] a10



excuse me for the error