Question

我最近开始使用gulp来保持我的开发项目井井有条，并且我遇到了一些我无法弄清楚的东西。所以这是我的任务：

gulp.task('jsassemble', function () { 
   return gulp
   .src('vendor/proj/**/**/src/assets/js/*.js')
   .pipe(concat('all.js'))
  .pipe(gulp.dest('public/js'));
});

如您所见，它将获取vendor / proj / anyFolder / anySubFolder / src / assets / js中的每个js文件，将它们放在一起，重命名新创建的js＆＃39; all.js＆＃39 ;然后把它放在public / js中。问题是我想保留文件夹层次结构，例如：

Source = vendor / proj / anyFolder1 / anySubFolder1 / src / assets / js / * .js
Destination = public / js / anyFolder1 / anySubFolder1 / src / assets / js / all.js

Source = vendor / proj / anyFolder1 / anySubFolder2 / src / assets / js / * .js
Destination = public / js / anyFolder1 / anySubFolder2 / src / assets / js / all.js

而不是简单地将这些文件夹中的所有内容都放入1并且只有public / js / all.js

反正有吗？我已经尝试过谷歌了，但是我无法用几句话正确地表达我的问题而且给出了不想要的结果：/

Answer 1

您可以创建保留文件夹层次结构的功能。在此页面（http://www.jamescrowley.co.uk/2014/02/17/using-gulp-packaging-files-by-folder/）中，您可以找到解决方案。

var fs = require('fs');
var path = require('path');
var es = require('event-stream');
var gulp = require('gulp');
var concat = require('gulp-concat');
var rename = require('gulp-rename');
var uglify = require('gulp-uglify');

var scriptsPath = './src/scripts/';

function getFolders(dir){
    return fs.readdirSync(dir)
      .filter(function(file){
        return fs.statSync(path.join(dir, file)).isDirectory();
      });
}

gulp.task('scripts', function() {
   var folders = getFolders(scriptsPath);

   var tasks = folders.map(function(folder) {
      return gulp.src(path.join(scriptsPath, folder, '/*.js'))
        .pipe(concat(folder + '.js'))
        .pipe(gulp.dest(scriptsPath))
        .pipe(uglify())
        .pipe(rename(folder + '.min.js'))
        .pipe(gulp.dest(scriptsPath));
   });

   return es.concat.apply(null, tasks);
});

Answer 2

感谢@caballerog让我走上了正确的道路，这里是解释的代码：

//get every folder from a 'pathTo/Something'
function getFolders(dir){
   return fs.readdirSync(dir)
       return fs.statSync(path.join(dir, file)).isDirectory();

}


var projectsRoot = 'vendor/proj/';
var pathToJsFiles = '/src/assets/js/';
var pathToPublic = 'public/js/';

gulp.task('scripts', function() {

var sites = [];
var pathToProjects = [];

// Fetching every folders in vendor/proj
projects = getFolders(projectsRoot);

// Fetching every subfolder in vendor/proj/something
for(index in projects){
  sites.push(getFolders(projectsRoot + '/' + projects[index]));
}

// Pushing every projects/site that exists into an array
for(var i=0;i<projects.length;i++){
  for(var j=0; j<sites.length; j++)
    if(sites[i][j] != null)
      pathToProjects.push(projects[i] + '/' + sites[i][j]);
}

// Fetching every JS on vendor/proj/pathToAProject/pathToJsFiles
// concatenate them together
// and sending them to pathToPublic/pathToAProject/all.js
var tasks = pathToProjects.map(function(pathToAProject) {
   return gulp.src( projectsRoot + pathToAProject + pathToJsFiles + '/*.js')
    .pipe(concat('all.js'))
    .pipe(gulp.dest(pathToPublic + pathToAProject));
});

return es.concat.apply(null, tasks);
});

TL：DR =获取pathToPublic / someFolder / someFolder / PathToJS中的每个JS文件。

gulp-concat：从未知来源到同一目的地

2 个答案: