Using vimrc function to pass python arguments

Question

I am attempting to create a vimrc function that will be used in an autocmd. The function must simply call a python script and pass the file name as an argument.

.vimrc

fu! Test(filename)
    let filename = expand("%:t")
    "echom filename
    !test.py filename

example.py

#!usr/bin/python
import sys

print sys.argv[1]

If I uncomment the echo line, example.py is echo'd correctly. If I try to execute as it is displayed above, however, the string filename is passed literally.

Is there any way around this?

Nolen Royalty · Accepted Answer · 2016-07-28 21:13:37Z

1

Sure, you can use the execute command to execute a string, which is built from the command you want and the variable concatenated together:

fu! Test()
    let filename = expand("%:t")
    execute "!test.py " . l:filename
endfunction

I've omitted the filename argument in your Test function because it doesn't seem to be used

answered Jul 28, 2016 at 21:13

Nolen Royalty

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Comments

Meninx - メネンックス · Accepted Answer · 2016-07-28 21:14:35Z

1

You have two options either you pass the filename directly as an argument or pass it as a local variable:

fu! Test(filename)
    "echom a:filename
    execute "!test.py ".a:filename

or

fu! Test()
    let l:filename = expand("%:t")
    "echom filename
    execute "!test.py ". l:filename

answered Jul 28, 2016 at 21:14

Meninx - メネンックス

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