Browserify编译为完整的系统路径(使用watchify&gulp)

Question

我来自AMD,似乎已经做错了.

我做了这样的设置:

client/js/index.js (entry point)
client/js/test.js

我希望它建立:

build/app.js

index.js需要test.js像这样:

var test = require('./test');

我的gulp watchify任务看起来像这样:

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

// https://github.com/gulpjs/gulp/blob/master/docs/recipes/fast-browserify-builds-with-watchify.md
gulp.task('watch', function () {
    var bundler = watchify(browserify('./client/js/index.js', watchify.args));

    bundler.on('update', rebundle);

    function rebundle () {
        return bundler.bundle()
            // Log errors if they happen.
            .on('error', function(e) {
                gutil.log('Browserify Error', e.message);
            })
            .pipe(source('app.js'))
            .pipe(gulp.dest('./build'));
    }

    return rebundle();
});

编译后的代码看起来很糟糕,因为test.js我看到绝对的本地路径对于任何消费代码的人来说肯定是破坏或冗余的？

PS我正在运行没有args的任务(只是gulp watch)

(function e(t,n,r){function s(o,u){if(!n[o]){if(!t[o]){var a=typeof require=="function"&&require;if(!u&&a)return a(o,!0);if(i)return i(o,!0);var f=new Error("Cannot find module '"+o+"'");throw f.code="MODULE_NOT_FOUND",f}var l=n[o]={exports:{}};t[o][0].call(l.exports,function(e){var n=t[o][1][e];return s(n?n:e)},l,l.exports,e,t,n,r)}return n[o].exports}var i=typeof require=="function"&&require;for(var o=0;o<r.length;o++)s(r[o]);return s})({"./client/js/index.js":[function(require,module,exports){
var test = require('./test');

var ab = function(a, b2) {
    return a + b2;
};

module.exports = ab;
},{"./test":"/Users/dtobias/Sites/browserify-test/client/js/test.js"}],"/Users/dtobias/Sites/browserify-test/client/js/test.js":[function(require,module,exports){
var helloworld = function () {
    console.log('hello world');
};

module.exports = helloworld;
},{}]},{},["./client/js/index.js"]);

Answer 1

watchify用于观察文件的更改并自动更新它们,它不用于部署,您看到的路径是watchify.args在此行上使用的结果watchify(browserify('./client/js/index.js', watchify.args));在传递给fullPaths: truebrowserify 的参数中指出,这是watchify的一部分.能够在每次更改文件时加快构建过程.我建议做的是让watchify任务专门用于观看和浏览构建过程.

这可以通过设置一些开关并在监视任务中将其设置为true(从而修改代码)来轻松完成.

像这样的东西:

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

// https://github.com/gulpjs/gulp/blob/master/docs/recipes/fast-browserify-builds-with-watchify.md
gulp.task('build', function(){
    browserifyfun(false);
});
gulp.task('watch', function () {
    browserifyfun(true);
});
function browserifyfun(watch){
    var b;

    if(watch){
        b = watchify(browserify('./client/js/index.js', watchify.args));
        b.on('update', rebundle(b));
    }else{
        b = browserify('./client/js/index.js');
    }

    function rebundle (bundler) {
        return bundler.bundle()
            // Log errors if they happen.
            .on('error', function(e) {
                gutil.log('Browserify Error', e.message);
            })
            .pipe(source('app.js'))
            .pipe(gulp.dest('./build'));
    }

    return rebundle(b);
}

从这里修改代码