gulp - wrap plugin (which uses through2) output with string - node.js

I would like to know how exactly can I manipulate the output of my Gulp plugin so, for example, no matter how many files are passed to the plugin, it will wrap the output with a string. Currently I cannot know when does the last file is done.
The super simplified example below will iterate on 3 files and will create a new file named output.js and in it there will be three times the string xxx (xxxxxxxxx).
I would like the plugin itself to wrap the contents so the output will
be: +xxxxxxxxx+.
How can I do this?
Thanks!
Gulpfile
var gulp = require('gulp');
var concat = require('gulp-concat');
var foo = require('./index');
gulp.task('default', function() {
gulp.src([a.html, b.html, c.html])
.pipe(foo())
.pipe(concat('output.js'))
.pipe(gulp.dest('./test/output'))
});
The most basic gulp plugin (index.js):
var through2 = require('through2'),
gutil = require('gulp-util');
var PLUGIN_NAME = 'foo';
module.exports = function( options ){
// through2.obj(fn) is a convenience wrapper around
// through2({ objectMode: true }, fn)
return through2.obj(function( file, enc, callback ){
file.contents = new Buffer( 'xxx' );
this.push(file);
callback();
});
}
I understand the files are currently simply returned modified, but what I don't understand is how to append text and return the concatenated result that I want, while keeping it OK with Gulp working standards.
The "real" plugin should actually wrap the files results with:
var foo = { FILES_CONTENT }
where FILES_CONTENT will actually be a a concatenated string of all the files:
"file_name" : "file_content",
"file_name" : "file_content",
...

I would make the following changes to your gulpfile.js:
var gulp = require('gulp');
var foo = require('./index.js');
gulp.task('default', function() {
return gulp.src(['a.html', 'b.html', 'c.html'])
.pipe(foo({fileName:'output.js', varName:'bar'}))
.pipe(gulp.dest('./test/output'))
});
Since your foo() plugin itself will concatenate all the files, there's no need to use gulp-concat at all. Instead your plugin should accept an option fileName that provides the name of the generated file. I've also added another option varName that will provide the name of the var in the output file.
I'll assume that a.html, b.html and c.html are simple HTML files, something like this:
<h1 class="header">a</h1>
As you've already realized you need to concat all the files in the plugin itself. That's not really difficult however and doesn't require a lot of code. Here's a index.js which does exactly that:
var through2 = require('through2'),
gutil = require('gulp-util'),
path = require('path'),
File = require('vinyl');
var PLUGIN_NAME = 'foo';
module.exports = function(options) {
var files = { };
var outputFile = null;
return through2.obj(function(file, enc, callback){
outputFile = outputFile || file;
var filePath = path.relative(file.base, file.path);
files[filePath] = file.contents.toString();
callback();
}, function(callback) {
outputFile = outputFile ? outputFile.clone() : new File();
outputFile.path = path.resolve(outputFile.base, options.fileName);
outputFile.contents = new Buffer(
'var ' + options.varName + ' = ' +
JSON.stringify(files, null, 2) + ';'
);
this.push(outputFile);
callback();
});
}
Since you want to output a key/value mapping from file names to file contents our transformFunction just stores both of those things in a regular JavaScript object files. None of the input files themselves are emitted. Their names and contents are just stored until we have all of them.
The only tricky part is making sure that we respect the .base property of each file as is customary for gulp plugins. This allows the user to provide a custom base folder using the base option in gulp.src().
Once all files have been processed through2 calls the flushFunction. In there we create our output file with the provided fileName (once again making sure we respect the .base property).
Creating the output file contents is then just a matter of serializing our files object using JSON.stringify() (which automatically takes care of any escaping that has to be done).
The resulting ./test/output/output.js will then look like this:
var bar = {
"a.html": "<h1 class=\"header\">a</h1>\n",
"b.html": "<h1 class=\"header\">b</h1>\n",
"c.html": "<h1 class=\"header\">c</h1>\n"
};

You should use the gulp pipeline technique (standard).
This means that you can use the gulp-insert package in order
to add the string xxx.
var insert = require('gulp-insert');
.pipe(insert.append('xxx')); // Appends 'xxx' to the contents of every file
You can also prepend, append and wrap with this package and it support of course the gulp standards.
So the full example will be:
var gulp = require('gulp');
var concat = require('gulp-concat');
var foo = require('./index');
var insert = require('gulp-insert');
gulp.task('default', function() {
gulp.src([a.html, b.html, c.html])
.pipe(foo()
.pipe(insert.append('xxx'))
.pipe(concat('output.js'))
.pipe(gulp.dest('./test/output'))
});

Related

Passing a variable between pipes in Gulp 3.9.1

Using gulp 3.9.1
I am attempting to return a bunch of files and perform a task that requires a var to be passed between two pipes.
I'm using node uuid to create a v3 UUID for each file path to
ultimately end up with a uuid for each page. I'm grabbing the file path with gulp-print.
I want to store that uuid value as a var. In the next pipe Im using
gulp-inject-string to write it into the page during the build.
Help: Either I need help getting the file path inside the gulp-inject-string pipe or I need to pass the var between the two different pipes. If I globally set a var with a default value outside the src it gets passed easily to the pipe(inject).
Super simplified code below:
// test code
var gulp = require('gulp');
var print = require('gulp-print');
var inject = require('gulp-inject-string');
var reload = browserSync.reload;
const uuidv3 = require('uuid/v3');
var uuid;
gulp.task('uuid', function() {
return gulp.src('**/*.html'])
// create uuid
.pipe(print(function(filepath) {
uuid = uuidv3(filepath, uuidv3.URL);
return "compiled: " + filepath + ' uuid: ' + uuid;
}))
// need to to add UUIDv3 to each page
.pipe(inject.before('</head>', '<meta name="dc.identifier" content="' + uuid + '">'))
.pipe(gulp.dest('/prod/./'))
.pipe(reload({ stream: true }));
});
It's worth noting that I need a cross platform way to get the file path starting in the root of the project and including forward slashes. The gulp(print) does this perfectly starting at the root of the project and ignoring anything upstream from that point. The format of the path is important because it's one half of the equation in creating the uuid and the uuid's must match on Mac or PC platforms.
examples:
/index.html
/dir1/file.html
/dir1/dir2/dir3/file.html
var gulp = require('gulp');
var print = require('gulp-print');
var inject = require('gulp-inject-string');
const uuidv3 = require('uuid/v3');
var tap = require('gulp-tap');
// you can declare here
var uuid;
gulp.task('pages', function() {
// or you can declare here
var uuid;
return gulp.src('**/*.html')
// bunch of stuff happens here involving templating/minifying
// create uuid
.pipe(print(function(filepath) {
// then set it here and use it further below
// it will be available
uuid = uuidv3(filepath, uuidv3.URL);
return "compiled: " + filepath + ' uuid: ' + uuid;
}))
// need to to add UUIDv3 to each page
//.pipe(inject.before('</head>', '<meta name="dc.identifier" content="' + uuid + '">\n'))
.pipe(tap(function(file, t) {
return t.through(inject.before('</head>', '<meta name="dc.identifier" content="' + uuid + '">\n');
})
.pipe(gulp.dest('/prod/./'))
.pipe(reload({stream:true}));
});
You are just creating a variable at a higher scope that you can set and refer to later. If you need a bunch of them create an array with filepath as an index. But I would try it first as just a simple value.
I solved the problem. It was an amateur mistake. I returned the statement where the var was set so the var was essentially killed. Updated code that allows the var to pass through the pipes.
var gulp = require('gulp');
var print = require('gulp-print');
var replace = require('gulp-replace');
const uuidv3 = require('uuid/v3');
var uuid;
gulp.task('build', function() {
return gulp.src('**/*.html')
// get a cross-platform filepath and create a uuid
.pipe(print(function(filepath) {
uuid = uuidv3(filepath, uuidv3.URL);
}))
// inject uuid
.pipe(replace('dc.identifier" content=""', function() {
return 'dc.identifier" content="' + uuid + '"';
}))
.pipe(gulp.dest('/prod/./'));
});
The var uuid passes through the pipes just fine now. This code creates a UUID based on a cross-platform file path and injects it into an empty dc.identifier meta tag.

gulp-concat throws exception 'file.isNull(), index.js:39 TypeError: file.isNull is not a function

I am attempting to compile react (.jsx), coffeescript (.coffee), and vanilla javascript (.js) files using gulp to package all of the resulting .js files into one file, app.js, which gets loaded into my index.html page. I am generating a stream for each compilation type and using merge-stream to collect the contents of the 3 feeder streams into a single stream, which I am passing to gulp-concat to create app.js.
I am getting an exception from gulp-concat, index.js, line 39, letting me know that 'file' is not a function. Here is my entire gulpfile.js, the reference to gulp-concat is near the bottom of this section.
var browserify = require('browserify');
var coffee = require('gulp-coffee');
var concat = require('gulp-concat');
var gulp = require('gulp');
var gutil = require('gulp-util');
var mergeStream = require('merge-stream');
var reactify = require('reactify');
var sass = require('gulp-sass');
var source = require('vinyl-source-stream');
gulp.task('javascript', function(){
// convert .jsx files to .js, collecting them in a stream
var b = browserify();
b.transform(reactify); // use the reactify transform
b.add('./jsx-transforms.js');
jsxStream = b.bundle();
if (gutil.isStream(jsxStream)) {
gutil.log("jsxStream is a stream");
} else {gulp-concatgulp
gutil.log("jsxStream is not a stream");
}
merged = mergeStream(jsxStream);
if (gutil.isStream(merged)) {
gutil.log("merged is a stream");
} else {
gutil.log("merged is not a stream");
}
// collect all .js files in a stream
jsStream = gulp.src(['./client/**/*.js','./common/**/*.js']);
if (gutil.isStream(jsStream)) {
gutil.log("jsStream is a stream");
} else {
gutil.log("jsStream is not a stream");
}
merged.add(jsStream);
// compile all .coffee file to .js, collect in a stream
coffeeStream = gulp.src(['./client/**/*.coffee','./common/**/*.coffee'])
.pipe(coffee({bare: true}).on('error', gutil.log));
if (gutil.isStream(coffeeStream)) {
gutil.log("coffeeStream is a stream");
} else {
gutil.log("coffeeStream is not a stream");
}
merged.add(coffeeStream);
// concatenate all of the .js files into ./build/app.js
merged
.pipe(concat('app.js'))
.pipe(gulp.dest('./build'));
});
gulp.task('styles', function() {
gulp.src('./client/assets/stylesheets/**/*.scss')
.pipe(sass().on('error', sass.logError))
.pipe(concat('app.css'))
.pipe(gulp.dest('./build'));
});
gulp.task('default', ['javascript', 'styles']);
I have used gulp-concat before, but never hit this problem before.
Gulp streams are a very particular sort of stream: they are Node streams in object mode containing vinyl file objects. If your stream comes from somewhere other than gulp.src(), such as from the browserify API, then you will have to first convert the stream into the sort that gulp can deal with.
There are two steps you need to take. First, convert your browserify bundle stream into a stream containing a vinyl file object with vinyl-source-stream (which you have required but not used).
var source = require('vinyl-source-stream');
...
var jsxStream = b.bundle()
.pipe(source('bundle.js'));
Now there is another catch. Vinyl streams may be in one of two modes: streaming mode or buffer mode. Vinyl-source-stream gives you a stream in streaming mode. Many Gulp plugins, including gulp-concat, only support buffer mode. The fix for this is easy: use vinyl-buffer.
var source = require('vinyl-source-stream');
var buffer = require('vinyl-buffer');
...
var jsxStream = b.bundle()
.pipe(source('bundle.js'))
.pipe(buffer());
Now you have something you can merge with your other streams and pipe to gulp-concat. For more detail, see this recipe.

Duplicate Gulp Vinyl Stream with PassThrough- `TypeError: Invalid non-string/buffer chunk`

In Node, I am trying to duplicate a gulp vinyl stream using Passthrough. I get TypeError: Invalid non-string/buffer chunk when attempting c = fileStream.pipe(b);
I suspect it may be because fileStream is a gulp vinyl stream.
var pass = require('stream').PassThrough;
function duplicateStream(fileStream) {
b = new pass();
c = fileStream.pipe(b);
return c;
}
If you need to clone a gulp stream, you can just use gulp-clone. This task will write all of the single JS files to the out directory, as well as a concatenated bundle.js in the same directory.
var gulp = require('gulp');
var concat = require('gulp-concat');
var clone = require('gulp-clone');
var merge = require('merge-stream');
gulp.task('default', function () {
var scripts = gulp.src('assets/**/*.js');
var bundle = scripts.pipe(clone())
.pipe(concat('bundle.js'));
// Merge the streams together, then write them to the out folder
return merge(scripts, bundle).pipe(gulp.dest('out'));
});
https://github.com/mariocasciaro/gulp-clone

How do you create a file from a string in Gulp?

In my gulpfile I have a version number in a string. I'd like to write the version number to a file. Is there a nice way to do this in Gulp, or should I be looking at more general NodeJS APIs?
If you'd like to do this in a gulp-like way, you can create a stream of "fake" vinyl files and call pipe per usual. Here's a function for creating the stream. "stream" is a core module, so you don't need to install anything:
const Vinyl = require('vinyl')
function string_src(filename, string) {
var src = require('stream').Readable({ objectMode: true })
src._read = function () {
this.push(new Vinyl({
cwd: "",
base: "",
path: filename,
contents: Buffer.from(string, 'utf-8')
}))
this.push(null)
}
return src
}
You can use it like this:
gulp.task('version', function () {
var pkg = require('package.json')
return string_src("version", pkg.version)
.pipe(gulp.dest('build/'))
})
It's pretty much a one-liner in node:
require('fs').writeFileSync('dist/version.txt', '1.2.3');
Or from package.json:
var pkg = require('./package.json');
var fs = require('fs');
fs.writeFileSync('dist/version.txt', 'Version: ' + pkg.version);
I'm using it to specify a build date in an easily-accessible file, so I use this code before the usual return gulp.src(...) in the build task:
require('fs').writeFileSync('dist/build-date.txt', new Date());
This can also be done with vinyl-source-stream. See this document in the gulp repository.
var gulp = require('gulp'),
source = require('vinyl-source-stream');
gulp.task('some-task', function() {
var stream = source('file.txt');
stream.end('some data');
stream.pipe(gulp.dest('output'));
});
According to the maintainer of Gulp, the preferred way to write a string to a file is using fs.writeFile with the task callback.
var fs = require('fs');
var gulp = require('gulp');
gulp.task('taskname', function(cb){
fs.writeFile('filename.txt', 'contents', cb);
});
Source: https://github.com/gulpjs/gulp/issues/332#issuecomment-36970935
You can also use gulp-file:
var gulp = require('gulp');
var file = require('gulp-file');
gulp.task('version', function () {
var pkg = require('package.json')
return gulp.src('src/**')
.pipe(file('version', pkg.version))
.pipe(gulp.dest('build/'))
});
or without using gulp.src():
gulp.task('version', function () {
var pkg = require('package.json')
return file('version', pkg.version, {src: true})
.pipe(gulp.dest('build/'))
});
The gulp-header package can be used to prefix files with header banners.
eg. This will inject a banner into the header of your javascript files.
var header = require('gulp-header');
var pkg = require('./package.json');
var banner = ['/**',
' * <%= pkg.name %> - <%= pkg.description %>',
' * #version v<%= pkg.version %>',
' * #link <%= pkg.homepage %>',
' * #license <%= pkg.license %>',
' */',
''].join('\n');
gulp.src('./foo/*.js')
.pipe(header(banner, { pkg: pkg } ))
.pipe(gulp.dest('./dist/')
Gulp is a streaming build system leveraging pipes.
If you simply want to write a new file with an arbitrary string, you can use built in node fs object.
Using the string-to-stream and vinyl-source-stream modules:
var str = require('string-to-stream');
var source = require('vinyl-source-stream');
var gulp = require('gulp');
str('1.4.27').pipe(source('version.txt')).pipe(gulp.dest('dist'));
Here's an answer that works in 2019.
Plugin:
var Vinyl = require('vinyl');
var through = require('through2');
var path = require('path');
// https://github.com/gulpjs/gulp/tree/master/docs/writing-a-plugin#modifying-file-content
function stringSrc(filename, string) {
/**
* #this {Transform}
*/
var transform = function(file, encoding, callback) {
if (path.basename(file.relative) === 'package.json') {
file.contents = Buffer.from(
JSON.stringify({
name: 'modified-package',
version: '1.0.0',
}),
);
}
// if you want to create multiple files, use this.push and provide empty callback() call instead
// this.push(file);
// callback();
callback(null, file);
};
return through.obj(transform);
}
And in your gulp pipeline:
gulp.src([
...
])
.pipe(stringSrc('version.json', '123'))
.pipe(gulp.dest(destinationPath))
From source: https://github.com/gulpjs/gulp/tree/master/docs/writing-a-plugin#modifying-file-content
The function parameter that you pass to through.obj() is a _transform
function which will operate on the input file. You may also provide an
optional _flush function if you need to emit a bit more data at the
end of the stream.
From within your transform function call this.push(file) 0 or more
times to pass along transformed/cloned files. You don't need to call
this.push(file) if you provide all output to the callback() function.
Call the callback function only when the current file (stream/buffer)
is completely consumed. If an error is encountered, pass it as the
first argument to the callback, otherwise set it to null. If you have
passed all output data to this.push() you can omit the second argument
to the callback.
Generally, a gulp plugin would update file.contents and then choose to
either:
call callback(null, file) or make one call to this.push(file)
This can also be achieved using gulp-tap
This can be especially helpful if you have identified multiple files that require this header. Here is relevant code (Also from gulp-tap documentation)
var gulp = require('gulp'),
tap = require('gulp-tap');
gulp.src("src/**")
.pipe(tap(function(file){
file.contents = Buffer.concat([
new Buffer('Some Version Header', 'utf8'),
file.contents
]);
}))
.pipe(gulp.dest('dist');

node.js require all files in a folder?

How do I require all files in a folder in node.js?
need something like:
files.forEach(function (v,k){
// require routes
require('./routes/'+v);
}};
When require is given the path of a folder, it'll look for an index.js file in that folder; if there is one, it uses that, and if there isn't, it fails.
It would probably make most sense (if you have control over the folder) to create an index.js file and then assign all the "modules" and then simply require that.
yourfile.js
var routes = require("./routes");
index.js
exports.something = require("./routes/something.js");
exports.others = require("./routes/others.js");
If you don't know the filenames you should write some kind of loader.
Working example of a loader:
var normalizedPath = require("path").join(__dirname, "routes");
require("fs").readdirSync(normalizedPath).forEach(function(file) {
require("./routes/" + file);
});
// Continue application logic here
I recommend using glob to accomplish that task.
var glob = require( 'glob' )
, path = require( 'path' );
glob.sync( './routes/**/*.js' ).forEach( function( file ) {
require( path.resolve( file ) );
});
Base on #tbranyen's solution, I create an index.js file that load arbitrary javascripts under current folder as part of the exports.
// Load `*.js` under current directory as properties
// i.e., `User.js` will become `exports['User']` or `exports.User`
require('fs').readdirSync(__dirname + '/').forEach(function(file) {
if (file.match(/\.js$/) !== null && file !== 'index.js') {
var name = file.replace('.js', '');
exports[name] = require('./' + file);
}
});
Then you can require this directory from any where else.
Another option is to use the package require-dir which let's you do the following. It supports recursion as well.
var requireDir = require('require-dir');
var dir = requireDir('./path/to/dir');
I have a folder /fields full of files with a single class each, ex:
fields/Text.js -> Test class
fields/Checkbox.js -> Checkbox class
Drop this in fields/index.js to export each class:
var collectExports, fs, path,
__hasProp = {}.hasOwnProperty;
fs = require('fs');
path = require('path');
collectExports = function(file) {
var func, include, _results;
if (path.extname(file) === '.js' && file !== 'index.js') {
include = require('./' + file);
_results = [];
for (func in include) {
if (!__hasProp.call(include, func)) continue;
_results.push(exports[func] = include[func]);
}
return _results;
}
};
fs.readdirSync('./fields/').forEach(collectExports);
This makes the modules act more like they would in Python:
var text = new Fields.Text()
var checkbox = new Fields.Checkbox()
One more option is require-dir-all combining features from most popular packages.
Most popular require-dir does not have options to filter the files/dirs and does not have map function (see below), but uses small trick to find module's current path.
Second by popularity require-all has regexp filtering and preprocessing, but lacks relative path, so you need to use __dirname (this has pros and contras) like:
var libs = require('require-all')(__dirname + '/lib');
Mentioned here require-index is quite minimalistic.
With map you may do some preprocessing, like create objects and pass config values (assuming modules below exports constructors):
// Store config for each module in config object properties
// with property names corresponding to module names
var config = {
module1: { value: 'config1' },
module2: { value: 'config2' }
};
// Require all files in modules subdirectory
var modules = require('require-dir-all')(
'modules', // Directory to require
{ // Options
// function to be post-processed over exported object for each require'd module
map: function(reqModule) {
// create new object with corresponding config passed to constructor
reqModule.exports = new reqModule.exports( config[reqModule.name] );
}
}
);
// Now `modules` object holds not exported constructors,
// but objects constructed using values provided in `config`.
I know this question is 5+ years old, and the given answers are good, but I wanted something a bit more powerful for express, so i created the express-map2 package for npm. I was going to name it simply express-map, however the people at yahoo already have a package with that name, so i had to rename my package.
1. basic usage:
app.js (or whatever you call it)
var app = require('express'); // 1. include express
app.set('controllers',__dirname+'/controllers/');// 2. set path to your controllers.
require('express-map2')(app); // 3. patch map() into express
app.map({
'GET /':'test',
'GET /foo':'middleware.foo,test',
'GET /bar':'middleware.bar,test'// seperate your handlers with a comma.
});
controller usage:
//single function
module.exports = function(req,res){
};
//export an object with multiple functions.
module.exports = {
foo: function(req,res){
},
bar: function(req,res){
}
};
2. advanced usage, with prefixes:
app.map('/api/v1/books',{
'GET /': 'books.list', // GET /api/v1/books
'GET /:id': 'books.loadOne', // GET /api/v1/books/5
'DELETE /:id': 'books.delete', // DELETE /api/v1/books/5
'PUT /:id': 'books.update', // PUT /api/v1/books/5
'POST /': 'books.create' // POST /api/v1/books
});
As you can see, this saves a ton of time and makes the routing of your application dead simple to write, maintain, and understand. it supports all of the http verbs that express supports, as well as the special .all() method.
npm package: https://www.npmjs.com/package/express-map2
github repo: https://github.com/r3wt/express-map
Expanding on this glob solution. Do this if you want to import all modules from a directory into index.js and then import that index.js in another part of the application. Note that template literals aren't supported by the highlighting engine used by stackoverflow so the code might look strange here.
const glob = require("glob");
let allOfThem = {};
glob.sync(`${__dirname}/*.js`).forEach((file) => {
/* see note about this in example below */
allOfThem = { ...allOfThem, ...require(file) };
});
module.exports = allOfThem;
Full Example
Directory structure
globExample/example.js
globExample/foobars/index.js
globExample/foobars/unexpected.js
globExample/foobars/barit.js
globExample/foobars/fooit.js
globExample/example.js
const { foo, bar, keepit } = require('./foobars/index');
const longStyle = require('./foobars/index');
console.log(foo()); // foo ran
console.log(bar()); // bar ran
console.log(keepit()); // keepit ran unexpected
console.log(longStyle.foo()); // foo ran
console.log(longStyle.bar()); // bar ran
console.log(longStyle.keepit()); // keepit ran unexpected
globExample/foobars/index.js
const glob = require("glob");
/*
Note the following style also works with multiple exports per file (barit.js example)
but will overwrite if you have 2 exports with the same
name (unexpected.js and barit.js have a keepit function) in the files being imported. As a result, this method is best used when
your exporting one module per file and use the filename to easily identify what is in it.
Also Note: This ignores itself (index.js) by default to prevent infinite loop.
*/
let allOfThem = {};
glob.sync(`${__dirname}/*.js`).forEach((file) => {
allOfThem = { ...allOfThem, ...require(file) };
});
module.exports = allOfThem;
globExample/foobars/unexpected.js
exports.keepit = () => 'keepit ran unexpected';
globExample/foobars/barit.js
exports.bar = () => 'bar run';
exports.keepit = () => 'keepit ran';
globExample/foobars/fooit.js
exports.foo = () => 'foo ran';
From inside project with glob installed, run node example.js
$ node example.js
foo ran
bar run
keepit ran unexpected
foo ran
bar run
keepit ran unexpected
One module that I have been using for this exact use case is require-all.
It recursively requires all files in a given directory and its sub directories as long they don't match the excludeDirs property.
It also allows specifying a file filter and how to derive the keys of the returned hash from the filenames.
Require all files from routes folder and apply as middleware. No external modules needed.
// require
const { readdirSync } = require("fs");
// apply as middleware
readdirSync("./routes").map((r) => app.use("/api", require("./routes/" + r)));
I'm using node modules copy-to module to create a single file to require all the files in our NodeJS-based system.
The code for our utility file looks like this:
/**
* Module dependencies.
*/
var copy = require('copy-to');
copy(require('./module1'))
.and(require('./module2'))
.and(require('./module3'))
.to(module.exports);
In all of the files, most functions are written as exports, like so:
exports.function1 = function () { // function contents };
exports.function2 = function () { // function contents };
exports.function3 = function () { // function contents };
So, then to use any function from a file, you just call:
var utility = require('./utility');
var response = utility.function2(); // or whatever the name of the function is
Can use : https://www.npmjs.com/package/require-file-directory
Require selected files with name only or all files.
No need of absoulute path.
Easy to understand and use.
Using this function you can require a whole dir.
const GetAllModules = ( dirname ) => {
if ( dirname ) {
let dirItems = require( "fs" ).readdirSync( dirname );
return dirItems.reduce( ( acc, value, index ) => {
if ( PATH.extname( value ) == ".js" && value.toLowerCase() != "index.js" ) {
let moduleName = value.replace( /.js/g, '' );
acc[ moduleName ] = require( `${dirname}/${moduleName}` );
}
return acc;
}, {} );
}
}
// calling this function.
let dirModules = GetAllModules(__dirname);
Create an index.js file in your folder with this code :
const fs = require('fs')
const files = fs.readdirSync('./routes')
for (const file of files) {
require('./'+file)
}
And after that you can simply load all the folder with require("./routes")
If you include all files of *.js in directory example ("app/lib/*.js"):
In directory app/lib
example.js:
module.exports = function (example) { }
example-2.js:
module.exports = function (example2) { }
In directory app create index.js
index.js:
module.exports = require('./app/lib');

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