Search pattern between two quotes and make 1 row

Shell Programming and Scripting
1

Hi All,

My file

cat file
" test1 test1 "
" test1 test1
test1 test1"
"test1 test1
test1 test1
test1 test1 "

How to achieve this
i want the result:

cat file
test1 test1
test1 test1 test1 test1
test1 test1 test1 test1 test1 test1
2

Any attempts / ideas / thoughts from your side?

3

No sir in shell scripts, but im working on notepad++ with regular expression

find: [^"]*"([^"]+)"[^"]*
replace: \1

I can get the pattern between the two quotes but I cant make it in one line.

4

One option:

awk '{while (!/"$/) {getline X; $0 = $0 " " X}; gsub (/ *" */, _)} 1' file
test1 test1
test1 test1 test1 test1
test1 test1 test1 test1 test1 test1

or

sed  '{:L; N; $!bL;}; s/\n/ /g; s/ *\(" *\)\+/\n/g; s/^\n\|\n$//g; ' file
test1 test1
test1 test1 test1 test1
test1 test1 test1 test1 test1 test1
1 Like
5

Thanks a lot Sir, This is working