Making a copy of all .acignore->.gitignore in place, recursively

Question

I have the following file structure as an example

    c
    -Git
     -GitBranchTest
         .acignore
         [lots more files]
      -subfolder
         .acignore
         [lots more files]

I am trying to make a copy of all the .acignore files and rename to .gitignore, so when I am done it should look like this

    c
    -Git
     -GitBranchTest
         .acignore
         .gitignore
         [lots more files]
      -subfolder
         .acignore  
         .gitignore  
         [lots more files]

What I have tried and the error I got:

I hope this is clear enough. Please help.

score 1 · Accepted Answer · answered Aug 14 '20 at 15:27

1

This should do the trick:

find ./ -name '.acignore' | sed 's/.acignore//g' | xargs -I {} cp {}.acignore {}.gitignore

It's "./" if you are inside the "c" directory. If you are above it's "./c"

answered Aug 14 '20 at 15:27

Lorem ipsum

892
5
15

Thank you. I found another way to do it that is way more verbose. I still very new to bash shell scripting. Ive posted mine below, but will mark yours as the answer. – n4rzul Aug 14 '20 at 15:33
Thx for upvoting :). Xargs is basically a foreach loop. That is why mine is one line. – Lorem ipsum Aug 14 '20 at 15:40

score 0 · Answer 2 · answered Aug 14 '20 at 15:34

My attempt that works, but is very verbose :)

for i in  $(find /c/Git/GitBranchTest -name ".*" -print | grep acignore);do         
        from=$i
        replacement=.gitignore
        to=${from/.acignore/$replacement}
       echo "cp $from $to" | sh
done

Making a copy of all .acignore->.gitignore in place, recursively

2 Answers2