Can a gulp plugin take in one file and return many files?

Question

I am attempting to wrap the RequireJS Optimizer in a gulp plugin. There have been other attempts to provide this functionality in gulp but none of them meet my needs.

A common approach to plugin development is to take files in, process them, and then hand them out; however that approach is not really compatible with how the RequireJS Optimizer works.

While it does not meet the gulp plugin guidelines, I am working on a different approach of taking in the optimizer build configuration, processing it, and handing out the resulting file(s).

// Read in the r.js configuration
gulp.src( './build.js' )
    // Optimize the files according to the config
    .pipe( optimize() )
    // Return the resulting files for additional processing
    .pipe( size() );

Can a gulp plugin take in one file and return many files? If so, how?

Answer 1

I ended up using vinyl-fs to pass along the output of the optimization process.

var requirejs   = require( 'requirejs' ),
    through     = require( 'through2' ),
    vfs         = require( 'vinyl-fs' );

return through.obj( function( file, encoding, done ){

// Parse config from file

requirejs.optimize( config,

    // Success callback
    function( buildResponse ){
        vfs.src( config.dir + '/**/*.js' )
            .on( 'data', this.push.bind( this ) )
            .on( 'error', this.emit.bind( this, 'error' ) )
            .on( 'end', done );
    }.bind( this ),

    // Error callback
    function( buildError ){
        // Handle error
    } );

Usages ends up looking like...

// Read in build configuration
gulp.src( 'build.js' )
    // Run files from the config through the optimizer
    .pipe( optimize( settings ) )
    // Measure the size of the optimized files
    .pipe( size() );