[SOLVED] How can I remove trailing characters from a string using shell script?

How can I remove trailing characters from a string using shell script?

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

Solution

To answer the first line of your question which asks to "remove the last n characters from a string", you can use the substring extraction feature in Bash:

A="123456"
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/,*$//'.

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

ssl01:49188,ssl999999:49188

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

X="ssl01:49188,ssl999999:49188,,,,,"
shopt -s extglob
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.