how to load multiple files into separate buffers from zsh command line?

Question

I have a list of files like step-1.py, step-2.py etc and if I try to open them with vim files-*.py then I get a list of buffers like:

:ls
  1 %a   "step-10.py"
  2      "step-11.py"
  3      "step-12.py"
  4      "step-13.py"
  5      "step-14.py"
  6      "step-15.py"
  7      "step-16.py"
  8      "step-17.py"
  9      "step-18.py"
 10      "step-19.py"
 11      "step-1.py" 
 12      "step-2.py" 
 13      "step-3.py" 
 14      "step-4.py" 
 15      "step-5.py" 
 16      "step-6.py" 
 17      "step-7.py" 
 18      "step-8.py" 
 19      "step-9.py"

but I want the files to be loaded in order so that the buffer number corresponds to the step number (:b6 should load a buffer with step-6.py file in it).

So I've been trying to use craft the list in the proper order and supply it to vim... like...

ls step-* | sort -n -t- -k2 | vim -

...but that just loads a single unnamed buffer with the contents step-1.py step-2.py step-3.py step-4.py step-5.py step-6.py step-7.py step-8.py step-9.py step-10.py step-11.py step-12.py step-13.py step-14.py step-15.py step-16.py step-17.py step-18.py step-19.py .

If I instead try that like files=$(ls step-* | sort -n -t- -k2); vim $files then that still just gives me an empty buffer like:

:ls
  1 %a   "step-1.py step-2.py step-3.py step-4.py step-5.py step-6.py step-7.py step-8.py step-9.py step-10.py step-11.py step-12.py step-13.py step-14.py step-15.py step-16.py step-17.py step-18.py step-19.py " line 1

How do I achieve this?

Is it an option to rename the first 9 files with a leading zero? e.g. step-01.py That would solve the problem. — Heptite, Aug 15 '21 at 00:36

Matt · Answer 1 · 2021-08-15T11:24:17.367

3

That should be enough:

ls -v *.py | xargs gvim

edited Aug 15 '21 at 11:24

answered Aug 15 '21 at 07:42

Matt

20,685
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1

Historically this invocation caused vim to complain about stdin. Is that no longer the case? – D. Ben Knoble Aug 15 '21 at 10:39
2

@D.BenKnoble It still does. Works well for Neovim or GVim. – Matt Aug 15 '21 at 11:25
xargs! of course. thanks for the reminder :) – alec Aug 15 '21 at 13:18

alec · Answer 2 · 2021-08-16T22:45:32.557

Cleanest way: vi $(ls step-* | sort -n -t- -k2).

In trying the approach described in the question, vim needs the filenames provided to it as an array. So take files=$(ls step-* | sort -n -t- -k2); vim $files and merely add ( and ) around the existing variable definition so that it looks like...

files=($(ls step-* | sort -n -t- -k2)); vim $files and vim will open that list of files in their respective buffers in the proper order:

:ls
  1 %a   "step-1.py"                    line 9
  2      "step-2.py"                    line 0
  3      "step-3.py"                    line 0
  4      "step-4.py"                    line 0
  5      "step-5.py"                    line 0
  6      "step-6.py"                    line 0
  7      "step-7.py"                    line 0
  8      "step-8.py"                    line 0
  9      "step-9.py"                    line 0
 10      "step-10.py"                   line 0
 11      "step-11.py"                   line 0
 12      "step-12.py"                   line 0
 13      "step-13.py"                   line 0
 14      "step-14.py"                   line 0
 15      "step-15.py"                   line 0
 16      "step-16.py"                   line 0
 17      "step-17.py"                   line 0
 18      "step-18.py"                   line 0
 19      "step-19.py"                   line 0

That explains it; zsh doesn’t word-split unquoted variables like sh and bash, and it does expand arrays without extra syntax (again, unlike bash). Please include mention of zsh in the question, as the problem and solution appear very differently in different shells. — D. Ben Knoble, Aug 15 '21 at 12:56
oh... if I go into bash my solution doesn't seem to work there... only the first buffer is loaded. — alec, Aug 15 '21 at 13:22

how to load multiple files into separate buffers from zsh command line?

2 Answers2