[英]How can I bundle js files in my lib with main app.js
现在,仅捆绑了我的app.js
和其中使用的文件。 我希望我的库中的文件也一起捆绑到相同的js文件中。 这是我的js文件夹中的文件夹结构:
.
├── app.js
├── components
└── libs
└── materialize.min.js
这是我的gulpfile,将它们捆绑在一起:
import gulp from 'gulp'
import source from 'vinyl-source-stream'
import buffer from 'vinyl-buffer'
import browserify from 'browserify'
import babelify from 'babelify'
import uglify from 'gulp-uglify'
import watchify from 'watchify'
const jsDirs = {
src: './client/js/app.js',
dest: './dist/js'
}
function buildBundle(b) {
b.bundle()
.pipe(source('bundle.js'))
.pipe(gulp.dest(jsDirs.dest))
}
gulp.task('scripts', () => {
let b = browserify({
cache: {},
packageCache: {},
fullPaths: true
})
b = watchify(b.transform(babelify))
b.on('update', () => buildBundle(b))
b.add(jsDirs.src)
buildBundle(b)
})
gulp.task('default', ['scripts'])
有什么办法可以包含app.js
未使用的libs js文件吗?
您应该能够多次调用b.require(path)
。 (这就是我为我做的事情):
import gulp from 'gulp'
import source from 'vinyl-source-stream'
import buffer from 'vinyl-buffer'
import browserify from 'browserify'
import babelify from 'babelify'
import uglify from 'gulp-uglify'
import watchify from 'watchify'
const jsDirs = {
src: './client/js/app.js',
dest: './dist/js',
requires: [
'./libs/materialize.min.js',
['./libs/whatever.js', 'whatever']
]
}
function buildBundle(b) {
b.bundle()
.pipe(source('bundle.js'))
.pipe(gulp.dest(jsDirs.dest))
}
gulp.task('scripts', () => {
let b = browserify({
cache: {},
packageCache: {},
fullPaths: true
});
b = watchify(b.transform(babelify))
[].concat(jsDirs.requires || []).forEach(function (req) {
var path = req,
expose = null;
if (typeof path !== 'string') {
expose = path[1];
path = path[0]
}
b.require(path, expose ? { expose: expose } : {});
});
b.on('update', () => buildBundle(b))
b.add(jsDirs.src)
buildBundle(b)
})
gulp.task('default', ['scripts'])
这也让您可以将lib暴露给未来
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