我有一个库lib.js，我想从中创建lib/a.js并且lib/b.js能够从脚本中client.js使用它var a = require('lib/a.js');，并且当我lib.js之前只包含已编译的库时它可以工作client.js（因此，lib.js必须声明一个require知道的函数lib/a.js）
我想我必须使用external，alias但我不确定什么是正确的方法
另外，是否有可能有一个 Gulp 文件为我的库中的文件夹自动创建所有别名？例如。为目录中的所有文件创建别名lib/？
这里有几个 gulp 任务可以帮助你分别构建你的通用 lib.js 和 client.js 包。
请注意，捆绑lib.js时必须将browserify告知b.require() lib/*.js，并且必须告知b.external()捆绑client.js时将单独加载的库
var path = require('path');
var gulp = require('gulp');
var browserify = require('browserify');
var concat = require('gulp-concat');
var transform = require('vinyl-transform');
gulp.task('build-lib', function () {
// use `vinyl-transform` to wrap around the regular ReadableStream returned by b.bundle();
// so that we can use it down a vinyl pipeline as a vinyl file object.
// `vinyl-transform` takes care of creating both streaming and buffered vinyl file objects.
var browserified = transform(function(filename) {
// basename, for eg: 'a.js'
var basename = path.basename(filename);
// define the exposed name that your client.js would use to require();
// for eg: require('lib/a.js'); // -> exposed name should be 'lib/a.js'
var expose = 'lib/' + basename;
return browserify(filename)
.require(filename, { expose: expose})
.bundle();
});
return gulp.src(['./lib/*.js'])
.pipe(browserified)
.pipe(concat('lib.js'))
.pipe(gulp.dest('./dist'));
});
gulp.task('build-client', function () {
var browserified = transform(function(filename) {
// filename = './client.js'
// let browserify know that lib/a.js and and lib/b.js are external files
// and will be loaded externally (in your case, by loading the bundled lib.js
// for eg: <script src='dist/lib.js'>)
return browserify(filename)
.external('lib/a.js')
.external('lib/b.js')
.bundle();
});
return gulp.src(['./client.js'])
.pipe(browserified)
.pipe(gulp.dest('./dist'));
});
gulp.task('default', ['build-lib', 'build-client']);