require-dir-object

Turn a directory into a tree of objects with the required version of each file.

Example

Given a directory of JavaScript files:

.
├── fish.js
├── folder_with_index
│   ├── index.js
│   └── other_file.js
├── index.js
├── other_file.js
└── other_folder
    ├── badger.js
    └── badger_fish.js

index.js

module.exports = require("require-dir-object")(__dirname, {case: "camel"});

Would be equivalent to:

module.exports = {
    fish: require("./fish"),
    otherFile: require("./other_file"),
    otherFolder: {
        badger: require("./other_folder/badger"),
        badgerFish: require("./other_folder/badger_fish")
    },
    folderWithIndex: require("./folder_with_index")
};

Options

| Name | Description | Type | Example | Default | | -------------:|:-------------------------------- |:-----------------------:|:------------------------|:-------------------| | case | Converts file naming method | string | {case: "camel"} | null | | depth | Limit sub-directory search depth | int | {depth: 3} | Number.MAX_VALUE | | exclude | Exclude files at a specific path | string or array | {exclude: "fish.js"} | [] | | ext | Search for a specific extension | string | {ext: ".xml"} | ".js" | | func | Call a function on found file | function | {ext: fs.readFileSync}| require |

Available case conversions are camel, snake, kebab, capitalize.

Notes

• Every folder is turned into a sub-object.
• If a folder has an index it will load that in favor of folder contents.

License

MIT