問題描述

我的 gulpfile.js 中有以下內容:

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(???));
});

我想將縮小的 css 文件輸出到以下路徑:

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

我是否誤解了如何使用來源/目的地?還是我試圖在一項任務中完成太多?

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

更新了對應站點目錄的輸出路徑.

Updated output paths to corresponding site directories.

推薦答案

我猜 為每個文件夾運行任務 配方可能會有所幫助.

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

更新

遵循配方中的想法，并為了給出想法而過度簡化您的示例，這可能是一個解決方案:

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);
});

這篇關于我可以使用具有多個源和多個目標的 Gulp 任務嗎?的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！