CODEWITHSUNDEEP
linuxbashshellunixscripting
mebb
mebb

Reputation: 163

file path manipulation and archive it dynamically using shellscript

Generated output but it is growing list.

i have the files generated as above but i need to manipulate it so it can be used following usecase.

1) Need to move all files from repo to corresponding v folder (delete repo folder but need to push its contents level up).

output need to be like

2) zip each output in seperate files named after its product and another zip with all contents.

basic script for task 1

for i in `ls -d -1 "$PWD/"**/`
do
cd $i && cd ..
mv repo/* .
rm -rf repo
done

it does its work but it looks ugly.

i am trying to get zip files generated based on product name. how can i do it ?

Upvotes: 0

Views: 39

Answers (1)

Paul Hodges
Paul Hodges

Reputation: 15273

First, simplify what you have a bit.

for i in "$PWD"/**/; do # this will only be directory names
  cd "$i"/.. && mv repo/* . && rm -rf repo
done

but I think we can refine it.

base=/mnt/data/workspace/deploy/package
for p in npm atom iterm vscode; do
  cd "$base/$p/" && mv repo/* . && rm -rf repo
done
cd "$base/"
zip package *
for p in npm atom iterm vscode; do zip "$p" "$p/"; done

Is that doing what you wanted?

Upvotes: 1

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