awk - 如何用 CLI 上某些行的逗号替换\n

Question

我有以下文本文件。

 # This is a test 1
    "watch"
    "autoconf"

    # This is another line 2
    "coreutils"
    "binutils"
    "screen"

    # This is another line 3
    "bash"
    "emacs"
    "nano"
    "bison"

    #  This is another line 4
    "libressl"
    "python"
    "rsync"
    "unzip"
    "vim"

我想将其更改为以下内容:

# This is a test 1
watch, autoconf

# This is another line 2
coreutils, binutils, screen

# This is another line 3
bash, emacs, nano, bison

#  This is another line 4
libressl, python, rsync, unzip, vim

删除前导空格，删除引号，用逗号替换新行。

到目前为止我得到了这个:

$ cat in.txt | sed 's/"//g' | sed 's/^[[:space:]]*//'> out.txt

# This is a test 1
watch 
autoconf

# This is another line 2
coreutils
binutils
screen

# This is another line 3
bash 
emacs
nano
bison
...

我不确定如何用逗号替换新行。我尝试了以下方法。

# no change
$ cat in.txt | sed 's/"//g' | sed 's/^[[:space:]]*//'| sed  's/\n/,/g'> out.txt

# changed all new lines
$ cat in.txt | sed 's/"//g' | sed 's/^[[:space:]]*//'| sed -z 's/\n/,/g'> out.txt
$ cat out.txt
# This is a test 1,watch,autoconf,,# This is another line 2,coreutils,binutils,screen,,# This is another line 3,bash,emacs,nano,bison,,#  This is another line 4,libressl,python,rsync,unzip,vim

我怎样才能做到这一点？

Answer 1

这可能对你有用(GNU sed):

sed -E 's/^\s*//;/^"(\S*)"/{s//\1/;H;$!d};x;s/.//;s/\n/, /gp;$d;z;x' file

去除所有行前面的空白。

去掉双引号并将这些词附加到保留空间。

否则，切换到保留空间，删除第一个引入的换行符，用 , 替换所有其他换行符，打印结果，然后切换回模式空间并打印。