How to use gulp webpack-stream to generate a proper named file?(如何使用 gulp webpack-stream 生成正确的命名文件?)
问题描述
目前我们使用
我不能使用 c2212af8f732662acc64.js
我需要将其命名为 bundle.js 或其他正常名称.
我们的 Webpack 配置:
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模块.exports = {条目:./entry.js",开发工具:源地图",输出: {devtoolLineToLine:真,sourceMapFilename: "app/assets/js/bundle.js.map",路径信息:真,路径:__dirname,文件名:产品?app/assets/js/bundle.min.js":app/assets/js/bundle.js"},模块: {装载机:[{测试:/.css$/,加载器:style!css"}]},插件:产品?[新的 webpack.optimize.UglifyJsPlugin({minimize: true})]:[]};
啊,我继续阅读 并想通了:
gulp.task('webpack', function() {返回 gulp.src('entry.js').pipe(webpack( 需要('./webpack.config.js') )).pipe(gulp.dest('app/assets/js'));});
^ 在这里我可以传入我的实际 webpack.config,它会使用我已经在那里设置的路径.在我的情况下,我刚刚删除了 app/assets/js
因为我现在 gulp 中有该路径.
但仍然没有世俗的想法,为什么在我创建的第一个任务中,它会生成随机哈希文件名?
Currently we're using Webpack for our Module loader, and Gulp for everything else (sass -> css, and the dev/production build process)
I want to wrap the webpack stuff into gulp, so all I have to do is type gulp
and it starts, watches and runs webpack and the rest of what our gulp is setup to do.
So I found webpack-stream and implemented it.
gulp.task('webpack', function() {
return gulp.src('entry.js')
.pipe(webpack({
watch: true,
module: {
loaders: [
{ test: /.css$/, 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:
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})
] : []
};
Ah I read on a bit further and figured it out:
gulp.task('webpack', function() {
return gulp.src('entry.js')
.pipe(webpack( require('./webpack.config.js') ))
.pipe(gulp.dest('app/assets/js'));
});
^ here I can just pass in my actual webpack.config and it will use the paths I have already set in there. In my case I just removed app/assets/js
since I have that path in now gulp instead.
Still no earthly idea though, why with the first task I created, it generates random hash filenames?
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本文标题为:如何使用 gulp webpack-stream 生成正确的命名文件?
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