Replace CR / LF in a text file only after a specific column

Question

I have a large text file that I would like to put in my ebook reader, but the formatting becomes wrong, because all the lines are rigidly wrapped in column 80 or in front of column 80 with CR / LF, and paragraphs / headings are not marked differently , there is also only one CR / LF.

I would like to replace all CR / LF after column 75 with a space. This will make most paragraphs continuous. (Not an ideal solution, but much better read.)

Is it possible to do this with regex? Preferably (linux) perl or sed oneliner, alternatively the regular expression Notepad ++.

Answer 1

perl -p -e 's/\s+$//; $_ .= length() <= 75 ? qq{\n} : q{ }' book.txt

Perl -p : , . -e. : , , , .

Answer 2

:

sed '/^$/! {:a;N;s/\(.\{75\}[^\n]*\)\n\(.\{75\}\)/\1 \2/;ta}' ebook.txt

, 75 .

Edit:

:

sed '/^.\{0,74\}$/ b; :a;N;s/\(.\{75\}[^\n]*\)\n\(.\{75\}\)/\1 \2/;ta; s/\n/ /g' ebook.txt

2:

/ ( 65, ), ( ):

sed 's/^.\{0,74\}$/&\n/' ebook.txt | fmt -w 65 | sed '/^$;s/\n//}'

DOS Unix, dos2unix :

dos2unix < ebook.txt | sed '/^.\{0,74\}$/ b; :a;N;s/\(.\{75\}[^\n]*\)\n\(.\{75\}\)/\1 \2/;ta; s/\n/ /g'

Answer 3

, vim, . V , .

:g/\%>74v$\n/j

Answer 4

cr/lf apperar lf cr, cr/lf. / .

regex 1: ^\r\n$ cr/lf's. . . cr/lf np ++.