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!
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.
Related
With the Browserify API and Gulp, I have this:
var browserify = require('browserify');
var gulp = require('gulp');
var source = require('vinyl-source-stream');
var dependencies = [
'lodash',
'./test.js',
];
gulp.task('lib', function() {
return browserify()
.require(dependencies)
.bundle()
.pipe(source('lib.js'))
.pipe(gulp.dest('./'));
});
gulp.task('app', function() {
return browserify('./app.js')
.external(dependencies)
.bundle()
.pipe(source('bundle.js'))
.pipe(gulp.dest('./'));
});
And in app.js I have this:
var _ = require('lodash');
var test = require('./test.js');
The Lodash line works fine, but the ./test.js does not work. I get the error Error: Cannot find module '/test.js'.
How do I get this to work?
For some reason, the key differs between bundle.js and lib.js. In lib.js, the key for test.js is the full path (/Users/gary/Projects/browserify-test/test.js) whereas in bundle.js it's looking for a module with the key ./test.js. If I manually change the latter to be the same as the former, then it works.
I'm guessing that ultimately, Browserify doesn't support require on local files that are excluded from the same bundle.
browserify needs an absolute path to retrieve the file and it leaves that as the bundle key. The way to fix it is to use the expose option...
In your build..
var dependencies = [
'lodash',
{file: './test.js', expose: 'test'},
];
and in app.js...
var _ = require('lodash');
var test = require('test');
I'm using Gulp in a VS2015 project to run jscs on JavaScript files with the fix option set. The intention is to modify the same file that is read (viz., source and destination are the same).
var gulp = require('gulp');
var jscs = require('gulp-jscs');
var chmod = require('gulp-chmod');
var exec = require('gulp-exec');
var ourJsFiles = // an array of files and globbed paths
gulp.task('jscs', function (callback) {
ourJsFiles.forEach(function (fn) {
gulp.src(fn, { base: './' })
.pipe(jscs({
"preset": "google",
"maximumLineLength": 160,
"validateIndentation": 3,
"fix": true
}))
.pipe(gulp.dest('./'));
});
callback();
});
But I do not want to process any files that are read-only. Is there already a way to detect this in Gulp on Windows?
There is a plugin which allows you to work with subset of files: gulp-filter.
One of options is to pass filter function which will receive vinyl file object, so for e.g. you could use stat.mode property of that object which holds permissions and do something like:
var filter = require('gulp-filter');
...
var writableFiles = filter(function (file) {
//https://github.com/nodejs/node-v0.x-archive/issues/3045
var numericPermission = '0'+(e.stat.mode & parseInt('777', 8)).toString(8);
return numericPermission[1]==='6'
});
...
gulp.src(....)
.pipe(writableFiles)
We've been using gulp and browserify to create builds of our project whenever a js file changes. As the project has grown this process has become incredibly slow, from 200ms -> ~5s. The project has 69 directories, 173 files, and a max depth of 4 folders. We are applying a few transforms. Here's our build code.
var buildJS = function (entryPoint, name, cb) {
var browserify = require('browserify');
var uglify = require('gulp-uglify');
var source = require('vinyl-source-stream');
var buffer = require('vinyl-buffer');
var reactify = require('reactify');
var literalify = require('literalify');
var brfs = require('brfs');
browserify()
.require(entryPoint + '/' + name + '.jsx')
.transform({
global: true
}, reactify)
.transform({
global: true
}, brfs)
.transform({
global: true
}, literalify.configure(literalifyConfig))
.external(config.libs)
.bundle({
debug: config.DEV,
//detectGlobals : false
})
.on('error', handleError)
.pipe(source('bundle.js'))
.pipe(buffer())
.pipe(gulpIf(!config.DEV, uglify()))
.pipe(gulp.dest(config.buildPath + '/' + name))
.on('finish', cb)
};
Is this just normal behavior based on our project size? Or are we doing something wrong?
That's because browserify always recompile everything. You should be using incremental builds instead if you want good perfs.
I've developed a gulp plugin precisely for that: https://github.com/ngryman/gulp-bro.
I am trying to get any variation of hbsfy or browserify-handlebars to compile correctly using browserify. Compiling results in the handlebars.js(hbsfy) code outputting to my browser. I've tried just using the browserify command browserify -t hbsfy app.js > bundle.js but it doesn't change anything
I haven't the reputation to post images but basically this is the output:
var templater = require("handlebars/runtime").default.template;module.exports = templater(function (Handlebars,depth0,helpers,partials,data) { this.compilerInfo = [4,'>= 1.0.0']; helpers = this.merge(helpers, Handlebars.helpers); data = data || {}; var buffer = "", stack1, helper, functionType="function", escapeExpression=this.escapeExpression; buffer += "
Hello "; if (helper = helpers.name) { stack1 = helper.call(depth0, {hash:{},data:data}); } else { helper = (depth0 && depth0.name); stack1 = typeof helper === functionType ? helper.call(depth0, {hash:{},data:data}) : helper; } buffer += escapeExpression(stack1) + "
"; return buffer; });
My template (template.hbs) is simply <h1>Hello {{name}}</h1>
My gulpfile setup:
var gulp = require('gulp');
var livereload = require('gulp-livereload');
var browserify = require('gulp-browserify');
var hbsfy = require('browserify-handlebars');
//var hbsfy = require('hbsfy'); //this one shows up the same way
gulp.task('scripts', function() {
return gulp.src('./app/app.js')
.pipe(browserify({
transform: [hbsfy]
}))
.pipe(rename('bundle.js'))
.pipe(gulp.dest('./build/js'))
.pipe(connect.reload());
});
and my js file:
var Handlebars = require('hbsfy/runtime');
var $ = require('jquery'),
router = require('./router/routerDefault'),
template = require('./template.hbs');
$(document).ready(function(){
document.body.innerHTML = template({name: 'browserify'});
})
Does anyone have any experience on how to handle this? Any suggestions would be heplful!
The cause of this issue is redundant compiling. Listing a transform in both the packages.json and the gulpfile.js will perform it twice, I believe. In my packages.json, I now just use this 'node':
"browserify": {
"transform": [
"hbsfy"
]
},
This will compile your templates for you. Your gulpfile.js DOES NOT require this section:
.pipe(browserify({
transform: [hbsfy]
}))
You can use either one. My scripts gulp task now looks like this:
gulp.task('scripts', function() {
return browserify('./app/app.js')
.bundle()
.pipe(source('bundle.js'))
.pipe(gulp.dest('./build/js'))
.pipe(connect.reload());
});
I am experiencing something similar.
Just curious, what OS are you using? Seems to affect Mac but Windows seems OK.
I'm not entirely sure what's causing this but I stopped using gulp-browserify as it is now blacklisted.
I followed the suggestions from this blog post and it seems to solve the issue: http://viget.com/extend/gulp-browserify-starter-faq
The last bit is most relevant.
EDIT:
While using gulp-browserify, I would also check if you've listed your transforms in package.json. I think you may only need to specify transforms in one place (either in gulpfile as you have now or in package.json).
Consider the following two files:
config.json
{
"vendorFiles": [
"vendor/angular/angular.js",
"vendor/angular-ui-router/release/angular-ui-router.js",
"vendor/angular-ui-utils/modules/utils.js"
]
}
gulpfile.js
var gulp = require("gulp"),
concat = require("gulp-concat"),
config = require("./config");
gulp.task("build-vendor", function() {
gulp.src(config.vendorFiles)
.pipe(concat("vendor.js"))
.pipe(gulp.dest("build"));
});
How can I eliminate the need to specify vendor/ for each file in config.json? That file is one that is manually edited by other developers by hand, so I want to make it as hassle-free on them as possible.
Ideally I'd like my gulpfile.js to take care of adding that prefix (somehow), and for my config.json to look like this:
{
"vendorFiles": [
"angular/angular.js",
"angular-ui-router/release/angular-ui-router.js",
"angular-ui-utils/modules/utils.js"
]
}
There may be a better way with a Gulp specific solution, but this should work.
var gulp = require("gulp"),
concat = require("gulp-concat"),
config = require("./config");
gulp.task("build-vendor", function() {
gulp.src(config.vendorFiles.map(function(a) {return 'vendor/' + a}))
.pipe(concat("vendor.js"))
.pipe(gulp.dest("build"));
});
Demo:
http://jsfiddle.net/AK4tP/
Can't you just do
var gulp = require("gulp"),
concat = require("gulp-concat"),
config = require("./config");
gulp.task("build-vendor", function() {
gulp.src(config.vendorFiles, {root: 'vendor/'})
.pipe(concat("vendor.js"))
.pipe(gulp.dest("build"));
});
Gulp should accept root option in src() although it's not documented.