sed用单个空行替换多个空行

Question

I have written a sed script, which replaces multiple blank lines with a single one, but it is not working as supposed to. I will be thankful to everybody, who can explain me why. Please do not refer me to working examples, I am familiar with Google. I just want to understand how sed works.

The code is

sed ':a;/^\n*$/{N;ba};s/^\n\n*/\n/' input_file

so the logic is simple: when sed reads the line and it is either blank or has several newline symbols (this is /^*\\n$ condition), I tell sed to append the next line to pattern space. as soon as a non-blank line is found, the substitution s/^\\n\\n*/\\n/ is done.

Everything works fine except the cases when I have blank lines in the end of the file. These blanks are not replaced with a single blank and I do not understand why.

Any ideas?

Answer 1

The problem is sed getting EOF while reading next line.

Your command is getting completed at the time of reading next line using N . Because, While reading the next line, sed getting EOF . So, It won't process s/^\\n\\n*/\\n/ substitution. That is why, You are not able to remove sequence of empty lines which has appeared in end of file.

My solution is:

sed ':a; /^\n*$/{ s/\n//; N;  ba};' yourfile

Answer 2

Instead of \\n\\n* , you could use \\n+

By the way, this is what it would look like as a Perl one-liner:

perl -0777 -pe 's/\n+/\n/g' yourfile

Or, for the same result but fewer substitutions:

perl -0777 -pe 's/\n\n+/\n/g' yourfile

Or, even more magical, try this compact solution suggested by @hwnd:

perl -00 -pe ''