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.
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'
};
I'd like to do something like this:
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'
};
I'm wondering if Gulp exposes this functionality. Perhaps node itself allows this.
(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.)
Thanks!
Writing the First Gulp Task. All Gulp configuration goes in a file called gulpfile. js located at the root of the project. The pattern for writing tasks is that you first load a plugin you're about to use and then define a task that is based on that plugin.
The src() and dest() methods are exposed by gulp to interact with files on your computer. src() is given a glob to read from the file system and produces a Node stream. It locates all matching files and reads them into memory to pass through the stream.
In addition to making sure you have gulp globally installed, make sure your gulp file is named *gulpfile. js* and that is in the same directory as where you are running gulp . Then simply run gulp task1 .
Besides option.cwd
, you can also use process.chdir(yourDir)
it could be used anywhere in a gulpfile. e.g.
process.chdir(yourDir);
var gulp = require('gulp');
Make sure your gulp is up-to-date( > 3.8.10), this may not work in older gulp.
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'));
Or, well, you can just:
gulp.src(..., {cwd: __dirname})
gulp.dest(..., {cwd: __dirname})
Something like:
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);
};
Look here and here.
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