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I'm working on an npm package, let's call it foo, that has a few external dependencies. One such dependency, bar requires a build flag in order to work with my project. If I were to manually install the dependencies I would say:
npm install bar --bar-option=1
... # other deps
npm install foo
node script_that_uses_foo.js
I would like the dependencies of foo to be installed automatically with npm install foo. So I have a section in my package.json file that looks like this:
"dependencies" : {
"bar": "file:../../bar-0.1.0.tgz",
"baz": "*"
}
This works fine, except that bar is installed without --bar-option=1. How can I tell npm to pass this argument to the install script of bar? I've looked through the npm documentation and haven't found what I'm looking for.
Thanks for your help.
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To declare an optional dependency, add the <optional/> tag to the declared dependency in the plugin pom. xml . If any of the classes in the unavailable optional dependencies are used in the plugin a NoClassDefFoundError will be thrown.
Execute npm install someDependency --save-optional to install a package as an optional dependency. The installed package will be put into optionalDependencies . When you want to avoid installing optional dependencies, you can execute npm ci --no-optional (e.g. on CI tools like GitLab CI).
To add dependencies and devDependencies to a package. json file from the command line, you can install them in the root directory of your package using the --save-prod flag for dependencies (the default behavior of npm install ) or the --save-dev flag for devDependencies.
I know this is really old, but I believe that in this case, you can
npm install foo --bar-option=1
npm will pass bar-option through to all dependencies, e.g. "bar" when installing them.
alternatively, inside of foo's package.json, you could define a preinstall script that does "npm install bar --bar-option=1"
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