Reputation: 4511
Is there an option in npm (or other tool) to print all used licenses? I have a project and I want to make sure I don't use a library which is under a license I can't use.
EDIT: Found out that many developers don't include the license in the package.json, so I had to find out manually using "npm docs package-name"
Upvotes: 79
Views: 47877
Reputation: 390
The following commands output the licenses of your dependencies. Because npm view
queries npm, it it may take some time.
For npm:
npm ls | tail -n +2 | awk '{print $2}' | xargs -I {} -L 1 npm view {} license | awk '!seen[$0]++
For bun (+ npm):
bun pm ls | tail -n +2 | awk '{print $2}' | sed -e $'s/\x1b\[[0-9;]*m//g' | xargs -I {} -L 1 npm view {} license | awk '!seen[$0]++
bun pm ls
only lists the top-level dependencies, which is what I wanted, whereas npm seems to list all dependencies in node_modules
? The sed
is to strip bun's output of ANSI colors, to not trip up npm view
. The last awk
deduplicates the output. This works on macOS, but hasn't been tested on other systems.
It's possible to add any property you want to extract from each dependency's package.json
, for example the name
(see npm help view
for more). This will output multiple lines in ini/toml format per dependency, and the deduplication no longer makes sense. Add --json
to get JSON objects.
For example ... npm view {} name license
:
name = '@antfu/eslint-config'
license = 'MIT'
...
Upvotes: 0
Reputation: 11953
Quick and easy way to check:
npx license-checker --summary
Great source for more information: https://medium.com/@fokusman/the-easiest-way-to-check-all-your-npm-dependency-licenses-753075ef1d9d
Upvotes: 4
Reputation: 396
if you want to get all licenses from a directory or it subdirectories you can use NPM License Crawler https://www.npmjs.com/package/npm-license-crawler. It was the best solution for me
Upvotes: 1
Reputation: 3865
Yarn has a command for this as well. yarn licenses list
renders short output, yarn licenses generate-disclaimer
renders all the actual license text to stdout (suitable for disclaimers, as the option would imply).
If you want to omit devDependencies
:
NODE_ENV=production yarn licenses list
For my purposes, the following command got me close enough:
yarn licenses list | grep License | \
grep -vE 'MIT|ISC|WTFPL|BSD|Apache|Unlicense|CC-BY|Public Domain'`
Upvotes: 29
Reputation: 1990
You can try this it you are on a Linux based system:
npm list -g --depth=0 | awk '{print $2}' | xargs -i npm view {} | grep license
You will have something like this:
license: 'MIT',
license: 'MIT',
license: 'MIT',
license: 'BSD',
license: 'MIT',
license: 'MIT',
license: 'BSD-2-Clause',
license: 'MIT',
.....................
.....................
.....................
license: 'BSD-2-Clause',
Upvotes: 3
Reputation: 14355
If you're using Atom, there is npm-license-checker to get the licenses from package.json
.
Upvotes: 0
Reputation: 3078
Having just done this for a large project, I can say it turns out this process is more of a headache to automate fully than you might think. It's easy to get many of them with some of the tricks listed here, but NPM package licenses are not published consistently, and can appear
In addition, you sometimes have to read a licenses to tell which well-known open source license it corresponds to.
The best tool I know to do this, that (unlike some of the other answers here) covers all these cases is the licensecheck package: https://github.com/marcello3d/node-licensecheck
It looks at package.json as well as common license files, and does a signature match against known licenses, so it accurately recognizes more licenses automatically. It also "normalizes" licenses against the standard SPDX list of licenses (https://spdx.org/licenses/).
Finally, Licensecheck also lets you save any remaining packages you needed to manually verify in your own license.json file (since you can't count on an external maintainer to change their package).
Taken together, this is a pretty robust solution.
Upvotes: 11
Reputation: 9134
I had exactly the same requirement, and wrote a node module to do this. Shameless self promotion I know, but it is open source and hope it can help resolve your issue. Let me know if you have any issues or suggestions.
The difference over the other answers is that it does not just use the package.json license declaration, but looks for potential license information in license and readme files in the project.
You can install using npm install -g nlf
Upvotes: 75
Reputation: 2222
I liked the question, and took the time to write a nodejs script for it:
var npm = require('npm');
npm.load(process.config,function(err){
npm.list(function(err,deps){
var names = Object.keys(deps.dependencies);
for(var i in names){
var depen = deps.dependencies[names[i]];
console.log('Licenses for :',names[i]);
depen.licenses.forEach(function(license,i){
console.log('License #'+(i+1));
console.log('- Title:',license.type);
console.log('- Url:\t',license.url);
});
}
});
});
this will output each license name and url for each module,
NOTE: must be executed in project folder and npm must be installed (npm install npm -g
sounds overkill but this is the npm js lib)
Upvotes: 3
Reputation: 14953
cd {project}/node_modules
ls | sed 's/$/\/package.json/' | xargs grep '"license[s]*"' -A 3
Could use some improvement, but it works (at least on osx, should work on linux, no idea about windows). You should see something like:
grunt/package.json: "licenses": [
grunt/package.json- {
grunt/package.json- "type": "MIT",
grunt/package.json- "url": "http://github.com/gruntjs/grunt/blob/master/LICENSE-MIT"
--
grunt-contrib-concat/package.json: "licenses": [
grunt-contrib-concat/package.json- {
grunt-contrib-concat/package.json- "type": "MIT",
grunt-contrib-concat/package.json- "url": "https://github.com/gruntjs/grunt-contrib-concat/blob/master/LICENSE-MIT"
--
Update:
If you wish to see the name of all modules, even those nested inside other modules, the following works (cred to @robertklep, slightly modified to still work when inside the node_modules directory):
find * -name package.json | xargs grep '"license[s]*"' -A 3
Upvotes: 23