我可以使用具有多个来源和多个目的地的Gulp任务吗? [英] Can I use a Gulp task with multiple sources and multiple destinations?

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问题描述

在我的gulpfile.js中有以下内容:

  var sass_paths = [
'./httpdocs- site1 / media / sass / ** / *。scss',
'./httpdocs-site2/media/sass/**/*.scss',
'./httpdocs-site3/media/ sass / ** / *。scss'
];

gulp.task('sass',function(){
return gulp.src(sass_paths)
.pipe(sass({errLogToConsole:true}))
.pipe(autoprefixer('last 4 version'))
.pipe(minifyCSS({keepBreaks:true}))
.pipe(rename({suffix:'.min'}))
.pipe(gulp.dest(???));
});

我想将我缩小的CSS文件输出到以下路径:

  ./ httpdocs-site1 / media / css 
./httpdocs-site2/media/css
./httpdocs-site3/ media / css

我误解如何使用来源/目的地吗?编辑:更新输出路径到相应的网站目录。

解决方案

我猜测

更新



遵循配方中的想法,并简单介绍一下您的示例,这可能是一个解决方案:

  var gulp = require('gulp'),
path = require('path'),
merge = require('merge-stream');

var folders = ['httpdocs-site1','httpdocs-site2','httpdocs-site3'];
$ b $ gulp.task('default',function(){

var tasks = folders.map(function(element){
return gulp.src(element +'/media/sass/**/*.scss',{base:element +'/ media / sass'})
// ...其他步骤...
.pipe(gulp .dest(element +'/ media / css'));
});

return merge(tasks);
});




I have the following in my gulpfile.js:

   var sass_paths = [
        './httpdocs-site1/media/sass/**/*.scss',
        './httpdocs-site2/media/sass/**/*.scss',
        './httpdocs-site3/media/sass/**/*.scss'
    ];

gulp.task('sass', function() {
    return gulp.src(sass_paths)
        .pipe(sass({errLogToConsole: true}))
        .pipe(autoprefixer('last 4 version'))
        .pipe(minifyCSS({keepBreaks:true}))
        .pipe(rename({ suffix: '.min'}))
        .pipe(gulp.dest(???));
});

I'm wanting to output my minified css files to the following paths:

./httpdocs-site1/media/css
./httpdocs-site2/media/css
./httpdocs-site3/media/css

Am I misunderstanding how to use sources/destinations? Or am I trying to accomplish too much in a single task?

Edit: Updated output paths to corresponding site directories.

解决方案

I guess that the running tasks per folder recipe may help.

Update

Following the ideas in the recipe, and oversimplifying your sample just to give the idea, this can be a solution:

var gulp = require('gulp'),
    path = require('path'),
    merge = require('merge-stream');

var folders = ['httpdocs-site1', 'httpdocs-site2', 'httpdocs-site3'];

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

    var tasks = folders.map(function(element){
        return gulp.src(element + '/media/sass/**/*.scss', {base: element + '/media/sass'})
            // ... other steps ...
            .pipe(gulp.dest(element + '/media/css'));
    });

    return merge(tasks);
});

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