Bash脚本-Sed命令

Question

I created a script to add 100 space characters at the end of each line:

#!/bin/ksh
sed -i -e 's/\n/            - 100 spaces  -                         /' $1     

but it doesn't work and I think it is because of \\n . Any thoughts?

Answer 1

Sed processes the content of the lines without the newline. Your code never sees a newline, so it cannot replace it.

Match the end of the string:

sed -i -e 's/$/ - 100 spaces - /' $1 

Answer 2

Although Karoly has already pointed out the error in your script, you could also save yourself typing 100 spaces by using a condition and break to make a sort of loop

sed ':1;/ \{100\}$/!{s/$/ /;b1}' file

Will print 100 space at the end of the line

If there are already spaces at the end and you want to add 100

sed 's/$/1/;:1;/1 \{100\}$/!{s/$/ /;b1};s/1\( *\)$/\1/' file

Answer 3

Just a suggestion to avoid typing 100 spaces (although I'm sure that by this time, you already have!) - use perl:

perl -pe 's/$/" " x 100/e' file

As the other answers have stated, this matches the end of the line and replaces with 100 spaces, using the e modifier to allow you to write " " x 100 and let perl do the work for you.

As with sed, you can modify the file in-place using the -i switch - as with sed, I'd suggest using something like -i.bak to create a backup file file.bak .

Answer 4

Could be the \\n as that's undefined by POSIX and so implementation dependent across seds but you also say you want add spaces to the end of a line while your script is trying to replace the end of line with spaces.

In any case this is not a job for sed, just use [s]printf("%100s","") in awk to create a string of 100 blanks, eg:

$ echo "foo bar" | awk '{printf "%s%10s%s\n", $1, "", $2}'
foo          bar