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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
715
In this method, you have to use the rev command. The rev command is used to reverse the line of string characterwise. Here, the rev command will reverse the string, and then the -c option will remove the first character. After this, the rev command will reverse the string again and you will get your output.
To remove the last three characters from the string you can use string. Substring(Int32, Int32) and give it the starting index 0 and end index three less than the string length. It will get the substring before last three characters.
To access the last n characters of a string, we can use the parameter expansion syntax ${string: -n} in the Bash shell. -n is the number of characters we need to extract from the end of a string.
To answer the title of you question with specifies cutting last n character in a string, you can use the substring extraction feature in Bash.
me@home$ 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.
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