Gulp不会以watchify,browserify退出

Question

我想设置gulp能够做两件事:1)使用watchify来监控文件中的更新并使用browserify自动重建更改,以及2)进行一次ad-hoc构建并退出.

#1似乎工作正常,但我无法让#2工作.我输入gulp build终端,所有东西都捆绑得很好,但是gulp不会退出或退出; 它只是坐在那里,我没有带回命令行.

我究竟做错了什么？这是整个gulpfile:

'use strict';

var gulp = require('gulp');
var browserify = require('browserify');
var watchify = require('watchify');
var source = require('vinyl-source-stream');
var gutil = require('gulp-util');

var b = watchify(browserify({
  cache: {},
  packageCache: {},
  entries: ['./app/app.js'],
  debug: true,
  transform: ['reactify']
}));

b.on('log', gutil.log);

var bundle = function() {
  return b.bundle()
    .pipe(source('bundle.js'))
    .pipe(gulp.dest('./dist'));
};

gulp.task('watch', function() {
  b.on('update', bundle);
});

gulp.task('build', function() {
  bundle();
});

gulp.task('default', ['watch', 'build']);

这是我终端的输出:

[11:14:42] Using gulpfile ~/Web Dev/event-calendar/gulpfile.js
[11:14:42] Starting 'build'...
[11:14:42] Finished 'build' after 4.28 ms
[11:14:45] 1657755 bytes written (2.99 seconds)

Gulp在11:14:45之后仍然在日志后运行,并且没有跳回到终端.

Answer 1

.bundle()不应该在watchify包装器上调用.以下修正了一切:

'use strict';

var gulp = require('gulp');
var browserify = require('browserify');
var watchify = require('watchify');
var source = require('vinyl-source-stream');
var gutil = require('gulp-util');

var b = function() {
  return browserify({
    cache: {},
    packageCache: {},
    entries: ['./app/app.js'],
    debug: true,
    transform: ['reactify']
  });
};

var w = watchify(b());

w.on('log', gutil.log);

var bundle = function(pkg) {
  return pkg.bundle()
    .pipe(source('bundle.js'))
    .pipe(gulp.dest('./dist'));
};

gulp.task('watch', function() {
  bundle(w);
  w.on('update', bundle.bind(null, w));
});

gulp.task('build', bundle.bind(null, b()));

gulp.task('default', ['watch']);