How to create a HTML file with the same name as my pug file each time I save using gulp?
All the docs on https://pugjs.org/ explain how to return pug in console...
You need task runner or module bundler for that. So choose one -grunt/gulp/webpack. Consider webpack as newest one and width best functionality.
Here an example width gulp as moust easy to understand from my point of view.
First install npm packages for compiling pug and watch for changes - npm install --save gulp-pug gulp-watch.
Then create and config your gulpfile.js.
First import an npm modules
var pug = require('gulp-pug');
var watch = require('gulp-watch');
Then create compiling task
gulp.task('pug',function() {
return gulp.src('PATH_TO_TEMPLATES/*.jade')
.pipe(pug({
doctype: 'html',
pretty: false
}))
.pipe(gulp.dest('./src/main/webapp/'));
});
And then create watcher
gulp.task('watch', function () {
return watch('PATH_TO_TEMPLATES/*.jade', { ignoreInitial: false })
.pipe(gulp.dest('pug'));
});
And run gulp-watch from you console.
Here an example width gulp 4.0
install npm packages for compiling pug/jade by following command
npm install --save gulp-pug .
create script with name gulpfile.js
//gulpfile.js
var gulp = require("gulp"), pug = require('gulp-pug');
function pugToHtml(){
return gulp.src('src')
.pipe(pug({
pretty: true
}))
.pipe(gulp.dest('dest'));
}
exports.pugToHtml = pugToHtml;
We can run the above code using the following command in your file directory:
gulp pugToHtml
You can now do the same without requiring gulp-watch as well.
Require module:
var pug = require('gulp-pug');
Task example:
//create task
gulp.task('pug', function buildHTML(){
gulp.src('src/pre/*.pug')
.pipe(pug())
.pipe(gulp.dest('src/post'));
});
Watch:
//pug serve
gulp.task('watch', ['pug'], function() {
gulp.watch(['src/pug-pre/*.pug'], ['pug']);
});
You can then run gulp watch. You can also set watch to the default task so you only have to write gulp in Terminal, if you are not automating anything else:
// default task
gulp.task('default', ['watch']);
Inexperienced users want to "see" the app that I've created in Node.js, but they don't want to use the console. According to them, it's a good idea to install it, and with a simple click, in desktop, they could "see" it.
They want to run the Node.js app as a Windows program. That's all!
How can I do it? Should I create a batch file?
**SOLVED**
An .bat file, renamed as "appstart.bat"
cd C:\Users\MyUser\MyApp
npm start
With shortcut in desktop.
You can also create a shortcut and set "Target" to...
C:\Windows\System32\cmd.exe /k "node index.js"
... and set "Start In" to the directory index.js is in.
Because Node-Red is usually installed globally _
a Node-Red.bat file could simply be
cd \
Node-Red
then use convertico.com to change the Node-Red.png file to Node-Red.ico.
Slick Willy
The bat file should look like the below
cd folder directory
node file.js
Example:
cd users\me\documents\mynodeproject\
node index.js
When you click the bat file shortcut it will open the node command prompt and your console app will launch.
You can use the create-desktop-shortcuts node module:
npm install --save create-desktop-shortcuts
Run this code to create the shortcut:
function createShortcut () {
const path = require('path');
const createDesktopShortcut = require('create-desktop-shortcuts');
const pathToYourScript = path.join(process.cwd, 'your-script.js');
const options = {
name: 'My cool script',
filePath: 'node',
arguments: '"' + pathToYourScript + '"'
};
const shortcutsCreated = createDesktopShortcut({
windows: options,
linux: options
// OSX shortcuts do not support arguments in shortcuts
});
if (shortcutsCreated) {
console.log('Everything worked correctly!');
} else {
console.log('Could not create the icon or set its permissions (in Linux if "chmod" is set to true, or not set)');
}
}
createShortcut();
I am developing a Gulpfile. Can it be made to restart as soon as it changes? I am developing it in CoffeeScript. Can Gulp watch Gulpfile.coffee in order to restart when changes are saved?
You can create a task that will gulp.watch for gulpfile.js and simply spawn another gulp child_process.
var gulp = require('gulp'),
argv = require('yargs').argv, // for args parsing
spawn = require('child_process').spawn;
gulp.task('log', function() {
console.log('CSSs has been changed');
});
gulp.task('watching-task', function() {
gulp.watch('*.css', ['log']);
});
gulp.task('auto-reload', function() {
var p;
gulp.watch('gulpfile.js', spawnChildren);
spawnChildren();
function spawnChildren(e) {
// kill previous spawned process
if(p) { p.kill(); }
// `spawn` a child `gulp` process linked to the parent `stdio`
p = spawn('gulp', [argv.task], {stdio: 'inherit'});
}
});
I used yargs in order to accept the 'main task' to run once we need to restart. So in order to run this, you would call:
gulp auto-reload --task watching-task
And to test, call either touch gulpfile.js or touch a.css to see the logs.
I created gulper that is gulp.js cli wrapper to restart gulp on gulpfile change.
You can simply replace gulp with gulper.
$ gulper <task-name>
I use a small shell script for this purpose. This works on Windows as well.
Press Ctrl+C to stop the script.
// gulpfile.js
gulp.task('watch', function() {
gulp.watch('gulpfile.js', process.exit);
});
Bash shell script:
# watch.sh
while true; do
gulp watch;
done;
Windows version: watch.bat
#echo off
:label
cmd /c gulp watch
goto label
I was getting a bunch of EADDRINUSE errors with the solution in Caio Cunha's answer. My gulpfile opens a local webserver with connect and LiveReload. It appears the new gulp process briefly coexists with the old one before the older process is killed, so the ports are still in use by the soon-to-die process.
Here's a similar solution which gets around the coexistence problem, (based largely on this):
var gulp = require('gulp');
var spawn = require('child_process').spawn;
gulp.task('gulp-reload', function() {
spawn('gulp', ['watch'], {stdio: 'inherit'});
process.exit();
});
gulp.task('watch', function() {
gulp.watch('gulpfile.js', ['gulp-reload']);
});
That works fairly well, but has one rather serious side-effect: The last gulp process is disconnected from the terminal. So when gulp watch exits, an orphaned gulp process is still running. I haven't been able to work around that problem, the extra gulp process can be killed manually, or just save a syntax error to gulpfile.js.
I've been dealing with the same problem and the solution in my case was actually very simple. Two things.
npm install nodemon -g (or locally if you prefer)
run with cmd or create a script in packages like this:
"dev": "nodemon --watch gulpfile.js --exec gulp"
The just type npm run dev
--watch specifies the file to keep an eye on. --exec says execute next in line and gulp is your default task. Just pass in argument if you want non default task.
Hope it helps.
EDIT : Making it fancy ;)
Now while the first part should achieve what you were after, in my setup I've needed to add a bit more to make it really user friend. What I wanted was
First open the page.
Look for changes in gulpfile.js and restart gulp if there are any
Gulp it up so keep an eye on files, rebuild and hot reload
If you only do what I've said in the first part, it will open the page every time. To fix it, create a gulp task that will open the page. Like this :
gulp.task('open', function(){
return gulp
.src(config.buildDest + '/index.html')
.pipe(plugins.open({
uri: config.url
}));
Then in my main tasks I have :
gulp.task('default', ['dev-open']);
gulp.task('dev-open', function(done){
plugins.sequence('build', 'connect', 'open', 'watch', done);
});
gulp.task('dev', function(done){
plugins.sequence('build', 'connect', 'watch', done);
});
Then modifying your npm scripts to
"dev": "gulp open & nodemon --watch gulpfile.js --watch webpack.config.js --exec gulp dev"
Will give you exactly what you want. First open the page and then just keep live reloading. Btw for livereload I use the one that comes with connect which always uses the same port. Hope it works for you, enjoy!
Another solution for this is to refresh the require.cache.
var gulp = require('gulp');
var __filenameTasks = ['lint', 'css', 'jade'];
var watcher = gulp.watch(__filename).once('change', function(){
watcher.end(); // we haven't re-required the file yet
// so is the old watcher
delete require.cache[__filename];
require(__filename);
process.nextTick(function(){
gulp.start(__filenameTasks);
});
});
I know this is a very old question, but it's a top comment on Google, so still very relevant.
Here is an easier way, if your source gulpfile.js is in a different directory than the one in use. (That's important!) It uses the gulp modules gulp-newer and gulp-data.
var gulp = require('gulp' )
, data = require('gulp-data' )
, newer = require('gulp-newer' )
, child_process = require('child_process')
;
gulp.task( 'gulpfile.js' , function() {
return gulp.src( 'sources/gulpfile.js' ) // source
.pipe( newer( '.' ) ) // check
.pipe( gulp.dest( '.' ) ) // write
.pipe( data( function(file) { // reboot
console.log('gulpfile.js changed! Restarting gulp...') ;
var t , args = process.argv ;
while ( args.shift().substr(-4) !== 'gulp' ) { t=args; }
child_process.spawn( 'gulp' , args , { stdio: 'inherit' } ) ;
return process.exit() ;
} ) )
;
} ) ;
It works like this:
Trick 1: gulp-newer only executes the following pipes, if the source file is newer than the current one. This way we make sure, there's no reboot-loop.
The while loop removes everything before and including the gulp command from the command string, so we can pass through any arguments.
child_process.spawn spawns a new gulp process, piping input output and error to the parent.
Trick 2: process.exit kills the current process. However, the process will wait to die until the child process is finished.
There are many other ways of inserting the restart function into the pipes.
I just happen to use gulp-data in every of my gulpfiles anyway. Feel free to comment your own solution. :)
Here's another version of #CaioToOn's reload code that is more in line with normal Gulp task procedure. It also does not depend on yargs.
Require spawn and initilaize the process variable (yargs is not needed):
var spawn = require('child_process').spawn;
var p;
The default gulp task will be the spawner:
gulp.task('default', function() {
if(p) { p.kill(); }
// Note: The 'watch' is the new name of your normally-default gulp task. Substitute if needed.
p = spawn('gulp', ['watch'], {stdio: 'inherit'});
});
Your watch task was probably your default gulp task. Rename it to watch and add a gulp.watch()for watching your gulpfile and run the default task on changes:
gulp.task('watch', ['sass'], function () {
gulp.watch("scss/*.scss", ['sass']);
gulp.watch('gulpfile.js', ['default']);
});
Now, just run gulp and it will automatically reload if you change your gulpfile!
try this code (only win32 platform)
gulp.task('default', ['less', 'scripts', 'watch'], function(){
gulp.watch('./gulpfile.js').once('change' ,function(){
var p;
var childProcess = require('child_process');
if(process.platform === 'win32'){
if(p){
childProcess.exec('taskkill /PID' + p.id + ' /T /F', function(){});
p.kill();
}else{
p = childProcess.spawn(process.argv[0],[process.argv[1]],{stdio: 'inherit'});
}
}
});
});
A good solution for Windows, which also works well with Visual Studio task runner.
/// <binding ProjectOpened='auto-watchdog' />
const spawn = require('child-proc').spawn,
configPaths = ['Gulpconfig.js', 'bundleconfig.js'];
gulp.task('watchdog', function () {
// TODO: add other watches here
gulp.watch(configPaths, function () {
process.exit(0);
});
});
gulp.task('auto-watchdog', function () {
let p = null;
gulp.watch(configPaths, spawnChildren);
spawnChildren();
function spawnChildren() {
const args = ['watchdog', '--color'];
// kill previous spawned process
if (p) {
// You might want to trigger a build as well
args.unshift('build');
setTimeout(function () {
p.kill();
}, 1000);
}
// `spawn` a child `gulp` process linked to the parent `stdio`
p = spawn('gulp', args, { stdio: 'inherit' });
}
});
Main changes compared to other answers:
Uses child-proc because child_process fails on Windows.
The watchdog exits itself on changes of files because in Windows the gulp call is wrapped in a batch script. Killing the batch script wouldn't kill gulp itself causing multiple watches to be spawned over time.
Build on change: Usually a gulpfile change also warrants rebuilding the project.
Install nodemon globally: npm i -g nodemon
And add in your .bashrc (or .bash_profile or .profile) an alias:
alias gulp='nodemon --watch gulpfile.js --watch gulpfile.babel.js --quiet --exitcrash --exec gulp'
This will watch for file gulpfile.js and gulpfile.babel.js changes. (see Google)
P.S. This can be helpful for endless tasks (like watch) but not for single run tasks. I mean it uses watch so it will continue process even after gulp task is done. ;)
Here's a short version that's easy to understand that you can set as a default task so you just need to type "gulp":
gulp.task('watch', function() {
const restartingGulpProcessCmd = 'while true; do gulp watch2 --colors; done;';
const restartingGulpProcess = require('child_process').exec(restartingGulpProcessCmd);
restartingGulpProcess.stdout.pipe(process.stdout);
restartingGulpProcess.stderr.pipe(process.stderr);
});
gulp.task('watch2', function() {
gulp.watch(['config/**.js', 'webpack.config.js', './gulpfile.js'],
() => {
console.log('Config file changed. Quitting so gulp can be restarted.');
process.exit();
});
// Add your other watch and build commands here
}
gulp.task('default', ['watch']);
I spent a whole day trying to make this work on Windows / Gulp 4.0.2, and I (finally) made it...
I used some solutions from people on this page and from one other page. It's all there in the comments...
Any change in any function inside "allTasks" will take effect on gulpfile.js (or other watched files) save...
There are some useless comments and console.logs left, feel free to remove them... ;)
const { gulp, watch, src, dest, series, parallel } = require("gulp");
const spawn = require('child_process').spawn;
// This function contains all that is necessary: start server, watch files...
const allTasks = function (callback) {
console.log('==========');
console.log('========== STARTING THE GULP DEFAULT TASK...');
console.log('========== USE CTRL+C TO STOP THE TASK');
console.log('==========');
startServer();
// other functions (watchers) here
// *** Thanks to Sebazzz ***
// Stop all on gulpfile.js change
watch('gulpfile.js', function (callback) {
callback(); // avoid "task didn't complete" error
process.exit();
});
callback(); // avoid "task didn't complete" error
}
// Restart allTasks
// ********************************************
// CALL GULPDEFAULT WITH THE GULP DEFAULT TASK:
// export.default = gulpDefault
// ********************************************
const gulpDefault = function (callback) {
let p = null;
watch('gulpfile.js', spawnChildren);
// *** Thanks to Sphinxxx: ***
// New behavior in gulp v4: The watcher function (spawnChildren()) is passed a callback argument
// which must be called after spawnChildren() is done, or else the auto-reload task
// never goes back to watching for further changes (i.e.the reload only works once).
spawnChildren(callback);
function spawnChildren(callback) {
/*
// This didn't do anything for me, with or without the delay,
// so I left it there, but commented it out, together with the console.logs...
// kill previous spawned process
if (p) {
// You might want to trigger a build as well
//args.unshift('build');
setTimeout(function () {
console.log('========== p.pid before kill: ' + p.pid); // a random number
console.log('========== p before kill: ' + p); // [object Object]
p.kill();
console.log('========== p.pid after kill: ' + p.pid); // the same random number
console.log('========== p after kill: ' + p); // still [object Object]
}, 1000);
}
*/
// `spawn` a child `gulp` process linked to the parent `stdio`
// ['watch'] is the task that calls the main function (allTasks):
// exports.watch = allTasks;
p = spawn('gulp', ['watch'], { stdio: 'inherit', shell: true });
// *** Thanks to people from: ***
// https://stackoverflow.com/questions/27688804/how-do-i-debug-error-spawn-enoent-on-node-js
// Prevent Error: spawn ENOENT
// by passing "shell: true" to the spawn options
callback(); // callback called - thanks to Sphinxxx
}
}
exports.default = gulpDefault;
exports.watch = allTasks;
Install gulp-restart
npm install gulp-restart
This code will work for you.
var gulp = require('gulp');
var restart = require('gulp-restart');
gulp.task('watch', function() {
gulp.watch(['gulpfile.js'], restart);
})
it will restart gulp where you do changes on the gulpfile.js
I have a folder layout such that:
/
-- css/
-- js/
-- apps/
-- -- myFirstApp/
-- -- mySecondApp/
-- -- ...
Each of these are git submodules, and have a corresponding Gruntfile, package.json, etc. What I want to do is the same sequence of commands, but differ depending on the respective package.json.
My command list is this:
npm install
grunt dist
copy app/css/[fileName].css (from package.json) to css/
copy app/js/[fileName].js to js/
copy app/js/[fileName].html to /
Is there a plugin or something I'm overlooking that I can use with grunt to do this? I don't want to do it statically if at all possible -- I'd like to only have to update the submodule list for this to work.
I don't know of any pre-built Grunt task that will do this for you, but writing the task isn't so difficult. You'll need to pull in the Node fs module to deal with the filesystem and obviously there will be some other things to do... here's a general structure for it with some code and some TODO's:
var fs = require("fs"),
path = require("path");
module.exports = function ( grunt ) {
grunt.initConfig({
... // all of your other Grunt config
// options for our new task
copymodulefiles: {
all: {
rootDir: "apps/"
}
}
});
// Here's our custom task definition
grunt.registerMultiTask("copymodulefiles", "Copies files from sub-projects", function() {
var done = this.async(), // tell Grunt this is an async task
root = grunt.config(this.name + "." + this.target + ".rootDir"),
modules = fs.readdirSync(root);
modules.forEach(function(dirName) {
var pkg = fs.readFileSync(root + dirName + path.sep + "package.json", "utf8");
pkgJson = JSON.parse(pkg);
// TODO: find what you need in the pkgJson variable
// copy files from wherever to wherever
// You can write a file like so:
fs.writeFile(theFilenameToWrite, "Contents of the new file", function (err) {
// (check for errors!)
// log it?
grunt.log.ok("file written!");
});
});
// Determine when all are complete and call "done()" to tell Grunt everything's complete
// call Grunt's "done" method to signal task completion
done();
});
};
Try with grunt-shell i found it perfect and did similar tasks like what you are trying to do.
Have a look at my Gruntfile.js configuration what i have written to run shell commands:
shell: {
multiple: {
command: ['bower install',
'mv bower_components/** public/',
'rm -rf bower_components'
].join('&&')
}
}
So here i am running bower, then i am copying its components to public folder and after that i am deleting the bower_components folder. So i guess from here onwards you can customize this script as per your usage.
I'm using Grunt (task-based command line build tool for JavaScript projects) in my project. I've created a custom tag and I am wondering if it is possible to run a command into it.
To clarify, I'm trying to use Closure Templates and "the task" should call the jar file to pre-compile the Soy file to a javascript file.
I'm running this jar from command line, but I want to set it as a task.
Alternatively you could load in grunt plugins to help this:
grunt-shell example:
shell: {
make_directory: {
command: 'mkdir test'
}
}
or grunt-exec example:
exec: {
remove_logs: {
command: 'rm -f *.log'
},
list_files: {
command: 'ls -l **',
stdout: true
},
echo_grunt_version: {
command: function(grunt) { return 'echo ' + grunt.version; },
stdout: true
}
}
Check out grunt.util.spawn:
grunt.util.spawn({
cmd: 'rm',
args: ['-rf', '/tmp'],
}, function done() {
grunt.log.ok('/tmp deleted');
});
I've found a solution so I'd like to share with you.
I'm using grunt under node so, to call terminal commands you need to require 'child_process' module.
For example,
var myTerminal = require("child_process").exec,
commandToBeExecuted = "sh myCommand.sh";
myTerminal(commandToBeExecuted, function(error, stdout, stderr) {
if (!error) {
//do something
}
});
If you are using the latest grunt version (0.4.0rc7 at the time of this writing) both grunt-exec and grunt-shell fail (they don't seem to be updated to handle the latest grunt). On the other hand, child_process's exec is async, which is a hassle.
I ended up using Jake Trent's solution, and adding shelljs as a dev dependency on my project so I could just run tests easily and synchronously:
var shell = require('shelljs');
...
grunt.registerTask('jquery', "download jquery bundle", function() {
shell.exec('wget http://jqueryui.com/download/jquery-ui-1.7.3.custom.zip');
});
Guys are pointing child_process, but try to use execSync to see output..
grunt.registerTask('test', '', function () {
var exec = require('child_process').execSync;
var result = exec("phpunit -c phpunit.xml", { encoding: 'utf8' });
grunt.log.writeln(result);
});
For async shell commands working with Grunt 0.4.x use https://github.com/rma4ok/grunt-bg-shell.