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bash - output command that can install listed npm global packages

I know npm ls -g --depth=0 will produce output

npm ls -g --depth=0

C:UsersleonAppDataRoamingnpm
+-- create-react-app@1.5.2
+-- eslint@5.2.0
+-- prettier@1.13.7
`-- serve@9.4.0


I'm looking for a bash command that can output the following based on the above output:

npm i -g create-react-app
npm i -g eslint
npm i -g prettier
npm i -g serve


2 Answers
2

you can use the sed tool to handle this

sed

npm ls -g --depth=0 | sed -nE 's/^W+/npm i -g /p'


@version

s/@.*//;

W+

[^A-Za-z0-9_]*

-E

Here is a slight refinement of the answer posted by @0.sh

npm ls -g --depth=0 |
sed -n 's/^[^A-Za-z0-9_]*//;s/@.*$//p' |
xargs -n 1 npm i -g


Moving the code into xargs causes the generated commands to be executed, not just printed.

xargs

If you prefer, the extraction could be done with Awk or whatever instead. Even grep;

grep

npm ls -g --depth=0 |
grep -o '[A-Za-z][^@]*' |
xargs -n 1 npm i -g


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