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Suppose I have a file dir1/this_file.txt and dir2/this_file.txt, and I want to insert different boilerplate code not based on the file name or extension, but based on the directory in which the file is created. How can I do that?

-- Edit --

Additionally, what if the boilerplate code depends on the filename of the file being created?

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  • If my answer answers your Q, please consider tagging it as the Answer. Thanks.
    – RichieHH
    Feb 16, 2020 at 14:14

2 Answers 2

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As phils said obtain the directory name

(file-name-directory (or buffer-file-name ""))

and then open a boilerplate template in that directory is maybe what you want rather than configure the template code in Emacs itself. I knocked this up and it seems to work.

  (defun insert-boilerplate ()
    (let ((boiler-plate-file (expand-file-name ".boilerplate" (file-name-directory (or buffer-file-name "")))))
      (if (file-exists-p boiler-plate-file)
          (insert-file-contents boiler-plate-file)))
    nil)
  (add-to-list 'find-file-not-found-functions #'insert-boilerplate)

Simply create a .boilerplate file in the directory and it should be good to go.

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  • thanks. Always welcome. I'll edit my answer.
    – RichieHH
    Feb 16, 2020 at 11:25
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    Or just use (if (file-exists-p ".boilerplate") (insert-file-contents ".boilerplate")) which will do the same.
    – Stefan
    Feb 16, 2020 at 15:00
  • Heh. Indeed. It includes the full path. Noted.
    – RichieHH
    Feb 16, 2020 at 19:20
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You can obtain the directory with:

(file-name-directory (or buffer-file-name ""))
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  • 1
    You can also just use default-directory which will not always be identical, but for all practical purposes it is.
    – Stefan
    Feb 16, 2020 at 14:57
  • Heh. Yes, I obviously had a bit of a brain-fade there...
    – phils
    Feb 16, 2020 at 15:24

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