How to use gulp webpack-stream to generate a proper named file?

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Currently we're using Webpack for our Module loader, and Gulp for everything else (sass -> css, and the dev/production build process)

I want to wrap the webpack stuff into gulp, so all I have to do is type gulp and it starts, watches and runs webpack and the rest of what our gulp is setup to do.

So I found webpack-stream and implemented it.

gulp.task('webpack', function() {
    return gulp.src('entry.js')
        .pipe(webpack({
            watch: true,
            module: {
                loaders: [
                    { test: /\.css$/, loader: 'style!css' },
                ],
            },
        }))
        .pipe(gulp.dest('dist/bundle.js'));
});

The problem is that it generates a random character name for the .js file, how are we suppose to use that in our app?

From the github repo:

The above will compile src/entry.js into assets with webpack into dist/ with the output filename of [hash].js (webpack generated hash of the build).

How do you rename these files? Also the new gulp task generates a new file everytime I save an edit:

enter image description here

I can't use c2212af8f732662acc64.js I need it to be named bundle.js or something else normal.

Our Webpack config:

var webpack = require('webpack');
var PROD = JSON.parse(process.env.PROD_DEV || '0');
// http://stackoverflow.com/questions/25956937/how-to-build-minified-and-uncompressed-bundle-with-webpack

module.exports = {
    entry: "./entry.js",
    devtool: "source-map",
    output: {
        devtoolLineToLine: true,
        sourceMapFilename: "app/assets/js/bundle.js.map",
        pathinfo: true,
        path: __dirname,
        filename: PROD ? "app/assets/js/bundle.min.js" : "app/assets/js/bundle.js"
    },
    module: {
        loaders: [
            { test: /\.css$/, loader: "style!css" }
        ]
    },
    plugins: PROD ? [
        new webpack.optimize.UglifyJsPlugin({minimize: true})
    ] : []
};
4 Answers
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