sed - removing last comma from listed value after

问题:

I'm using sed to replace my file of new lines \n with ',' which works fine however, in my last item, I don't want the ,.

How can I remove this?

Example:

sed 's/\n/,/g' myfile.out > myfile.csv

Output:

1,2,3,4,5,6,

回答1:

Well you can use labels:

$ cat file
1
2
3
4
5
6
$ sed ':a;N;s/\n/,/;ba' file
1,2,3,4,5,6

You can also use paste command:

$ paste -sd, file
1,2,3,4,5,6

回答2:

^{Consider jaypal singh's paste solution, which is the most efficient and elegant.}

An awk alternative, which doesn't require reading the entire file into memory first:

awk '{ printf "%s%s", sep, $0; sep = "," }' myfile.out > myfile.csv

If the output should have a trailing newline (thanks, Ed Morton):

awk '{ printf "%s%s", sep, $0; sep = "," } END { printf "\n" }' myfile.out > myfile.csv

• For the first input line, sep, due to being an uninitialized variable, defaults to the empty string, effectively printing just $0, the input line.

• Setting sep to "," after the first print ensures that all remaining lines have a , prepended.

• END { printf "\n" } prints a trailing newline after all input lines have been processed. (print "" would work too, given that print appends the output record separator (ORS), which defaults to a newline).

The net effect is that , is only placed between input lines, so the output won't have a trailing comma.

回答3:

You could add a second s command after the first: sed -z 's/\n/,/g ; s/,$//. This removes a comma at the end. (The option -z is from gnu sed and I needed it to get the first s command working.)