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.

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);
});