gulp-concat :从未知来源到相同目的地

Question

我最近开始使用gulp来组织我的开发项目，但我遇到了一些我无法理解的事情。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'));
});

正如您所看到的，它将获取公共中的每个js文件，将它们放在一起，将新创建的js重命名为'all.js‘，然后将其放入vendor/proj/anyFolder/anySubFolder/src/assets/js，/js中。问题是我想要gulp来保持文件夹的层次结构，例如：

来源= vendor/proj/anyFolder1/anySubFolder1/src/assets/js/*.js

目的地= public/js/anyFolder1/anySubFolder1/src/assets/js/all.js

来源= vendor/proj/anyFolder1/anySubFolder2/src/assets/js/*.js

目的地= 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放到一个1中，并且只有public/js/all.js

有没有办法做到这一点？我尝试先在谷歌上搜索，但我不能用几个词来恰当地表达我的问题，得到了不想要的结果:/

Answer 1

你可以创建一个保持文件夹层次结构的函数。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，他让我走上了正确的道路，下面是解释代码：

//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。