How to pass value of pwd as variable in SED to replace variable in a script file Post: 302613641

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LEARN ABOUT X11R4

pwd

PWD(1)                                                             User Commands                                                            PWD(1)

NAME

       pwd - print name of current/working directory

SYNOPSIS

       pwd [OPTION]...

DESCRIPTION

       Print the full filename of the current working directory.

       -L, --logical
              use PWD from environment, even if it contains symlinks

       -P, --physical
              avoid all symlinks

       --help display this help and exit

       --version
              output version information and exit

       If no option is specified, -P is assumed.

       NOTE:  your shell may have its own version of pwd, which usually supersedes the version described here.  Please refer to your shell's docu-
       mentation for details about the options it supports.

AUTHOR

       Written by Jim Meyering.

REPORTING BUGS

       GNU coreutils online help: <http://www.gnu.org/software/coreutils/>
       Report pwd translation bugs to <http://translationproject.org/team/>

COPYRIGHT

       Copyright (C) 2017 Free Software Foundation, Inc.  License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>.
       This is free software: you are free to change and redistribute it.  There is NO WARRANTY, to the extent permitted by law.

SEE ALSO

       getcwd(3)

       Full documentation at: <http://www.gnu.org/software/coreutils/pwd>
       or available locally via: info '(coreutils) pwd invocation'

GNU coreutils 8.28                                                 January 2018                                                             PWD(1)

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How to replace a string with a variable in a file using sed?

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