javascript - 如何在没有 gulp/grunt/webpack 的情况下将 google-closure-compiler-js 用于 node.js 应用程序？

Question

docs没有任何单独使用它的例子，但他们确实是这样说的:

Unless you're using the Gulp or Webpack plugins, you'll need to specify code via flags. Both jsCode and externs accept an array containing objects in the form {src, path, sourceMap}. Using path, you can construct a virtual filesystem for use with ES6 or CommonJS imports—although for CommonJS, be sure to set processCommonJsModules: true.

我已经根据文档创建了一个“compile.js”文件:

const compile = require('google-closure-compiler-js').compile;
const flags = {
  jsCode: [{path: './server/server.js'}],
  processCommonJsModules: true
};
const out = compile(flags);
console.info(out.compiledCode); 

在我的“./server/server.js”文件中，我放了一个 console.log 但它没有输出。不知道从这里去哪里......

Answer 1

借自icidasset/quotes .

在我看来，该路径并不打算像您使用的那样使用。

引用:

Using path, you can construct a virtual filesystem for use with ES6 or CommonJS imports—although for CommonJS, be sure to set processCommonJsModules: true.

因此，您必须扩展自己的资源，当您走这条路时，webpack 和 gulp 必须为您做些什么。

files=['./server/server.js']
files.map(f => {
    const out = compile({
      jsCode: [{ src: f.content }],
      assumeFunctionWrapper: true,
      languageIn: 'ECMASCRIPT5'
    });
    return out;
}