Bash：使用sed或tr将换行符替换为“”

Question

I'm trying to format output in a way that inserts newline characters after each 'line', with lines denoted by double quotes (""). The quotes themselves are temporary and to be stripped in a later step.

Input:

"a",1,"aa""b",2,"bb"

Output:

a,1,aa
b,2,bb

I've tried:

sed 's/""/\n/'
sed 's/""/\/g'
tr '""' '\n'

But tr seems to replace every quote character and sed seems to insert \\n as text instead of a newline. What can I do to make this work?

Answer 1

echo '"a",1,"aa""b",2,"bb"' |awk -v RS='""' '{$1=$1} {gsub(/"/,"")}1'
a,1,aa
b,2,bb

or using sed :

 echo '"a",1,"aa""b",2,"bb"' |sed -e 's/""/\n/' -e 's/"//g'   # OR sed -e 's/""/\n/;s/"//g'
a,1,aa
b,2,bb

awk solution: Here the default record separator is changed from new line to "" . So awk will consider the EOL when it hits "" .

sed solution: Here first "" are converted into new line and second replacement is to remove " from each line.

Answer 2

neech@nicolaw.uk:~ $ cat file.txt
"a",1,"aa""b",2,"bb"
neech@nicolaw.uk:~ $ sed 's/""/\n/' file.txt | tr -d '"'
a,1,aa
b,2,bb

Answer 3

You seem to be dealing with POSIX sed , which does not have support for the \\n notation. Insert an actual new-line into the pattern, either:

sed 's/""/\
/'

Or:

sed 's/""/\'$'\n''/'

Eg:

sed 's/""/\
/' | tr -d \"

Output:

a,1,aa
b,2,bb

Answer 4

As suggested by George Vasiliou if you have perl you could use:

>  echo '"a",1,"aa""b",2,"bb"' | perl -pe 's/""/"\n"/g;s/"//g'

This avoids the non portable sed problem.

Or for a crappy hack version.

Replace the "" with another character and then use tr (since tr should work with \\n) to replace it with \\n instead then remove the single " after.

So you can get the "" replaced with newline like this:

sed 's/""/#/g' | tr '#' '\n'

Then the rest follows:

> echo '"a",1,"aa""b",2,"bb"'| sed 's/""/#/g' | tr '#' '\n' | sed 's/\"//g'