Question

How can I remove the last n characters from a particular string using shell script?

This is my input:

ssl01:49188,,,
ssl01:49188,
ssl01:49188,,,,,
ssl01:49188,ssl999999:49188,,,,,
ssl01:49188,abcf999:49188,,,,,

The output should be in the following format:

ssl01:49188
ssl01:49188
ssl01:49188
ssl01:49188,ssl999999:49188
ssl01:49188,abcf999:49188

Answer 1

To answer the title of you question with specifies cutting last n character in a string, you can use the substring extraction feature in Bash.

me@home$ A="123456"
me@home$ echo ${A:0:-2}  # remove last 2 chars
1234

However, based on your examples you appear to want to remove all trailing commas, in which case you could use sed 's/,*$//'.

me@home$ echo "ssl01:49188,ssl999999:49188,,,,," | sed 's/,*$//'
ssl01:49188,ssl999999:49188

or, for a purely Bash solution, you could use substring removal:

me@home$ X="ssl01:49188,ssl999999:49188,,,,,"
me@home$ shopt -s extglob
me@home$ echo ${X%%+(,)}
ssl01:49188,ssl999999:49188

I would use the sed approach if the transformation needs to be applied to a whole file, and the bash substring removal approach if the target string is already in a bash variable.