[SOLVED] What is logical AND in Sed/Vim?

Issue

I would like to get an AND -operator to the following Vim command

:command! -nargs=1 GrepSource :! grep <args> `sed -n 's/^source \(.*\)/\1/p' %`

such that I can use the command to find sources at

# source <path>
# . <path> 
source <path>
. <path>

I am not sure whether it is possible or not to get AND -operator to SED or Vim.

How can you add an AND -operator to the SED command?

Solution

You can just use Vim regular expressions to find the <path> name, instead of using sed:

:command! -nargs=1 GrepSource :exe printf("!grep '<args>' '%s'", matchstr(getline('.'), '^#\=\s*\%(source\|\.\)\s*\zs\f\+'))

I think that command should do what you want. You can use the next command to test what is happening ('exe' is changed to 'echo'):

:command! -nargs=1 GrepSourceTest :echo printf("!grep '<args>' '%s'", matchstr(getline('.'), '^#\=\s*\%(source\|\.\)\s*\zs\f\+'))

The command must be used on a line in one of the following forms:

# source <somefile>
# . <somefile>
source <somefile>
. <somefile>

Answered By - too much php