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Re: [Help-bash] How to print a string in the quote form not in the escap
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From: |
Dennis Williamson |
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Subject: |
Re: [Help-bash] How to print a string in the quote form not in the escaped form? |
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Date: |
Tue, 26 Apr 2016 08:27:03 -0500 |
On Apr 26, 2016 8:21 AM, "Peng Yu" <address@hidden> wrote:
>
> readability.
>
> On Tue, Apr 26, 2016 at 8:06 AM, Matthew Cengia <address@hidden> wrote:
> > No, there is not. It's escaped; why does it matter how it's escaped?
> >
> > On Tue, Apr 26, 2016 at 10:58 PM, Peng Yu <address@hidden> wrote:
> >>
> >> Hi, printf %q will print the string in the escaped form.
> >>
> >> ~$ printf '%q\n' "'"
> >> \'
> >>
> >> I the above example, I'd like to print it as "'" (including the double
> >> quote). Is there a way to do so? Thanks.
> >>
> >> --
> >> Regards,
> >> Peng
> >>
> >
> >
> >
> > --
> > Regards,
> > Matthew Cengia
>
>
>
> --
> Regards,
> Peng
>
The shell removes the quotes from your argument before it is passed to
printf. If you want to output quotes you will need to include them in your
format string.
printf '"%q"\n' ""