text processing - How to remove last character from grep output? -

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i have text file following content in (for example):

in first line "one", second"two " & " 3 " , also"four    ".  in second line nested "foo "bar" baz""zoo" patterns.

i tried had strings between pair of quotes , ended command:

grep -po '"\k[^"]+"' file

what command gave me following:

one" 2 "  3 " 4    " foo "  baz" zoo"

and what want above result desired output be:

one 2   3  4     foo   baz zoo

please me remove last " above grep output. don't want remove spaces output. don't have words expanded multiline. e.g:

... "foo "bar" ba z""zoo" ...

please, please don't suggest me can use multiple commands, know can. i'm ask if can grep , options alone?

this possible through below grep one-liner.

$ grep -op '"\k[^"]+(?="(?:[^"]*"[^"]*")*[^"]*$)' file 1 2   3  4     foo   baz zoo

another hacky 1 through pcre verb (*skip)(*f),

$ grep -op '[^"]+(?=(?:"[^"]*"[^"]*)*[^"]*$)(*skip)(*f)|[^"]+' file 1 2   3  4     foo   baz zoo

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