如何压缩多个文件夹在gulp中生成多个.zip文件？ [英] How to zip multiple folders generating multiple .zip files in gulp?

问题描述

我的文件夹结构如下所示：

   -  / projects 
  -  / projects / proj1 
 -  / projects / proj2 
  -  / projects / proj3 
  -  / zips 
  

对于项目（proj1，proj2和proj3）中的每个文件夹，我想压缩每个文件夹的内容并生成 proj1.zip ， proj2.zip 和 proj3.zip 位于 / zips


以下示例函数根据 proj1 文件夹生成单个zip文件



  zip = require（'gulp-zip'）; 
 gulp.task（'default'，function（）{
 return gulp.src（'./ projects / proj1 / *'）
 .pipe（zip（'proj1.zip'）） ）
 .pipe（gulp.dest（'./ zips'））; 
}）; 
  

但是我怎么能为项目中的每个文件夹执行这样的任务？我可以通过 gulp.src（'./projects/ *'）来压缩所有文件夹，但是那又如何？

解决方案

老问题我知道，而且我很新，所以这可能被认为是黑客攻击，但我一直在努力做你正在做的事情，并最终与此相关。 / p>



我的文件结构如下：



  proj 
 proj / dist 
 proj / dist / sub01 
 proj / dist / sub02 
 proj / dist / sub03 
 proj / zip 
  

我的任务最终如下所示：



  var gulp =要求（ 吞掉）; 
 var foreach = require（gulp-foreach）; 
 var zip = require（gulp-zip）; 
 $ b gulp.task（zip-dist，function（）{
 return gulp.src（./ dist / *）
 .pipe（foreach（function流文件）{
 var fileName = file.path.substr（file.path.lastIndexOf（/）+ 1）; 
 gulp.src（./ dist /+ fileName +/ ** / *）
 .pipe（zip（fileName +。zip））
 .pipe（gulp.dest（./zipped））; 
 
 return流; 
}））; 
}）; 
  

它抓取 ./ dist 作为其来源，然后将其管理为 gulp-foreach 。



gulp-foreach 查看每个项目，我使用普通的javascript substr（）来获取我存储为变量的当前项目的名称。

最后我设置了一个新的 src 使用存储的fileName var并将结果传递给 gulp-zip 再次使用存储的变量作为压缩文件的名称。

结果是这样的结构：

  proj 
 proj / dist 
 proj / dist / sub01 
 proj / dist / sub02 
 proj / dist / sub03 
 proj / zipped 
 proj / zipped / sub01.zip 
 proj / zipped / sub02.zip 
 proj / zipped / sub03.zip 
  

再次，我是一百万m我不是专家，但是这对我很有用，如果我明白这个问题可能对你有用，或者至少给你一些想法。

My folder structure looks like this:

- /projects
- /projects/proj1
- /projects/proj2
- /projects/proj3
- /zips

For each folder in projects (proj1, proj2 and proj3) I want to zip contents of each of those folders and generate proj1.zip, proj2.zip and proj3.zip in /zips folder.

Following example function generates single zip file from proj1 folder

zip = require('gulp-zip');
gulp.task('default', function () {
    return gulp.src('./projects/proj1/*')
        .pipe(zip('proj1.zip'))
        .pipe(gulp.dest('./zips'));
});

But how I can execute such task for each folder in projects? I can get all folders to zip by gulp.src('./projects/*') but what then?

解决方案

Old question I know and I am pretty new to gulp so this might be considered a hack but I have just been trying to do what you are after and I ended up with this.

My file structure is this:

proj
proj/dist
proj/dist/sub01
proj/dist/sub02
proj/dist/sub03
proj/zipped

And my task ended up like this:

var gulp = require("gulp");
var foreach = require("gulp-foreach");
var zip = require("gulp-zip");

gulp.task("zip-dist", function(){
   return gulp.src("./dist/*")
       .pipe(foreach(function(stream, file){
          var fileName = file.path.substr(file.path.lastIndexOf("/")+1);
          gulp.src("./dist/"+fileName+"/**/*")
              .pipe(zip(fileName+".zip"))
              .pipe(gulp.dest("./zipped"));

          return stream;
       }));
});

It grabs all the first level contents of ./dist as its source and then pipes it to gulp-foreach.

gulp-foreach looks at each item and I use a plain javascript substr() to get the name of the current item which I store as a variable.

Finally I set a new src using the stored fileName var and pipe the result to gulp-zip using the stored var again as the name of the zipped file.

The result is a structure that looks like this:

proj
proj/dist
proj/dist/sub01
proj/dist/sub02
proj/dist/sub03
proj/zipped
proj/zipped/sub01.zip
proj/zipped/sub02.zip
proj/zipped/sub03.zip

Again, I am a million miles from being an expert but this worked for me and if I understand the question might work for you as well or at least give you some ideas.

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