在 gulpfile.js 中设置工作目录?

2022-01-12 00:00:00 javascript gulp

有没有办法在 gulpfile 中设置 Gulp 的工作目录，这样我就可以从子目录运行 gulp 命令而不会遇到任何问题?我对此进行了搜索，但没有找到我要查找的内容.

Is there a way to set the working directory for Gulp within a gulpfile, so that I can run a gulp command from a subdirectory without running into any issues? I ran a search for this and didn't find what I was looking for.

为了澄清，我知道为我正在使用的文件添加前缀.然而，而不是这个 -

To clarify, I'm aware of adding a prefix to the files I'm using. However, instead of this - 

var gulp = require('gulp');
var jshint = require('gulp-jshint');
...

var paths = {
    js: [__dirname + 'app/*/*.js', __dirname + '!app/lib/**'],
    css: __dirname + 'app/*/*.styl',
    img: __dirname + 'app/img/*',
    index: __dirname + '*.html',
    dist: __dirname + 'dist'
};

我想做这样的事情:

var gulp = require('gulp');
var jshint = require('gulp-jshint');
...

gulp.cwd(__dirname); // This would be much easier to understand, and would make future edits a bit safer.

var paths = {
    js: ['app/*/*.js', '!app/lib/**'],
    css: 'app/*/*.styl',
    img: 'app/img/*',
    index: '*.html',
    dist: 'dist'
};

我想知道 Gulp 是否公开了这个功能.也许节点本身允许这样做.

I'm wondering if Gulp exposes this functionality. Perhaps node itself allows this.

(我意识到当我运行命令时可能有一种方法可以自己执行命令行，但我想将它包含在 gulp 文件中，特别是出于分发目的.我希望 gulp 的工作目录与gulpfile 所在的目录.)

(I realize that there is likely a way to do command line itself when I run the command, but I would like to include it in the gulp file, especially for distribution purposes. I want the working directory for gulp to match the directory in which the gulpfile resides.)

谢谢！

推荐答案

你应该使用 path.join 因为它会处理正确的斜线，并且按照该路径您可以添加一个快捷方式:

Instead of concatenating strings by yourself, you should be using path.join since it will take care of the proper slash, and following that path you can add a shorcut:

var path = require('path'),
    p    = function () {

    Array
        .prototype
        .unshift
        .call(arguments, __dirname);

    return path.join.apply(path, arguments);
};

console.log(p('a', 'b', 'c'));

或者，你可以:

gulp.src(..., {cwd: __dirname})
gulp.dest(..., {cwd: __dirname})

类似:

var src = function (globs, options) {

    options = options || {};
    options.cwd = __dirname;

    return gulp.src(globs, options);
};

var dest = function (folder, options) {

    options = options || {};
    options.cwd = __dirname;

    return gulp.dest(folder, options);
};

看这里和这里.

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