0

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弄清楚的事情。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'));
});

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，将新创建的 js 重命名为“all.js”，然后将其放入 public/js 中。问题是我想让 gulp 保持文件夹层次结构，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 = 供应商/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入 1 并且只有 public/js/all.js

有什么办法吗？我试图先用谷歌搜索它，但我无法用几句话正确表达我的问题，并且得到了不想要的结果：/

4

2 回答 2

0

感谢@caballerog 让我走上了正确的道路，这是解释的代码：

//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。

于 2015-12-17T01:00:02.387 回答
0

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);
});
于 2015-12-16T22:55:41.597 回答