I have an npm script:
"scripts": {
"start": "webpack-dev-server --inline --config build/webpack.myconfig.js"
}
I'd like to be able to run the command as follows to replace "myconfig" with something arbitrary:
npm run start myOtherConfig
and have it replace myconfig. Is there a way to do substitution like this?
I don't think so. But you can create as many custom commands as you want.
"scripts": {
"start": "webpack-dev-server --inline",
"myconfig": "webpack-dev-server --inline --config build/webpack.myconfig.js",
"myotherconfig": "webpack-dev-server --inline --config build/webpack.myotherconfig.js",
}
And just do:
npm run myconfig or npm run myotherconfig
Related
i Have a nodejs application and its working fine but in my structure i dont have a src folder i just have all other files and am using MVC but am having the error
here is my script
"scripts": {
"start": "babel-node server.js",
"dev": "nodemon --exec npm start",
"build-babel": "babel -d ./build ./server.js -s",
"build": "babel ./Myfolder --out-dir dist --copy-files",
"serve": "node dist/index.js",
"sequelize": "sequelize"
I'm trying to configure my script (in package.json) in such a way that I can run webpack in production or development mode via cli.
package.json
"scripts": {
"start": "webpack serve",
"build": "webpack --mode=production"
in my webpack.config.js file, I was expecting process.env.NODE_ENV to equal to whatever I set the mode to be. Instead, I keep getting undefined please how can I make it work
For the Dev Server
First, install webpack-dev-server with following command:
npm install webpack-dev-server
Add following line to the package.json script:
"scripts": {
"start": "webpack-dev-server --mode development"
}
For Production build
Add the following line to the scripts section of package.json:
"build": "node_modules/.bin/webpack --mode production"
The scripts section in package.json with both the dev and production build will look like this:
"scripts": {
"build": "node_modules/.bin/webpack --mode production",
"start": "webpack-dev-server --mode development"
}
I have recently updated gulp to 4+ version and then gulp does not start the watch tasks anymore.
What am i doing wrong? below here is the updated version of code gulp 3+ to gulp 4, I am using gulp.series as per gulp4 docs.
npm run build command would have previously copied over the assets folder to 2 other places, its not anymore starting the gulp tasks and gulp watch.
Given myGulpFile.js
var gulp = require('gulp');
var path = require('path');
gulp.task('copytask', function() {
gulp.src(['assets/**/*']).pipe(gulp.dest('buildpath/assets'))
.pipe(gulp.dest(path.resolve(__dirname, '../somepath/assets')));
});
gulp.task('watchtask', gulp.series( function watchtask(){
gulp.watch('assets/**/*', gulp.series('copytask'));
}));
gulp.task('default',gulp.series('watchtask', function (done) {
console.log("Gulp started now from webpack!!");
done();
}));
Given my package.json for webpack
"main": "webpack.config.js",
"scripts": {
"build": "webpack --progress -c",
"watch": "webpack -w --progress -c",
"dev": "webpack-dev-server --devtool eval --progress --colors --content-base build",
"deploy": "NODE_ENV=production webpack --config webpack.production.config.js"
},
and my webpack.config.js
var webpack = require('webpack');
var something ...;
require('./myGulpFile');
var config = {
...
}
module.exports = config;
any help regarding this is much appreciated!
EDIT: To further clarify my question, I want to run gulp by running just one of the webpack commands and automatically start the gulp tasks (this was happening in gulp3+ version using above procedure) but broke after updating to gulp4 version.
I think now you need to use gulp to execute the default gulp and gulp <name-of-task> to execute the considered task gulp, for example gulp copytask.
Its working now, I ended up changing package.json commands and adding && gulp -f myGulpFile.js to the end of the build,dev,deploy commands as below. This will start the gulp tasks right after the webpack commands and you only have to run one initial webpack command npm run build
"scripts": {
"build": "webpack --progress -c && gulp -f myGulpFile.js",
"watch": "webpack -w --progress -c",
"dev": "webpack-dev-server --devtool eval --progress --colors --content-base build && gulp -f myGulpFile.js",
"deploy": "NODE_ENV=production webpack --config webpack.production.config.js && gulp -f myGulpFile.js"
}, ```
I am trying to target multiple environments from local while executing React app.
1. Development
2. Staging
3. Production
I am also trying to test for offline mode in any of the environments. So, the scripts what I have configured is as follows:
"staging-server": "nodemon server.js --environment=staging",
"staging": "concurrently -k \"npm:staging-server\" \"NODE_ENV='staging' PORT=3003 react-scripts start\"",
"prod": "npm run build && forever server.js --environment=production"
I am able to fetch environment arg using args inside my Express, but my local ui app is still showing development only when I console for process.env.NODE_ENV. I am also trying to set NODE_ENV with same line for staging, but still no luck. PORT setting is working but, the app is running in 3000 and 3003 both ports.
How to get rid of this? I would like to understand the staging configuration as well.
As per the docs, we cannot override NODE_ENV, but there is a room to create our own custom variables starting with REACT_APP_. So i configured to look as below:
Reference: https://facebook.github.io/create-react-app/docs/adding-custom-environment-variables
"staging": "concurrently -k \"npm:staging-server\" \"cross-env REACT_APP_ENVIRONMENT='staging' PORT=3003 react-scripts start\"",
And inside my UI application, I can fetch its value by consoling it like this:
console.log('REACT_APP_ENVIRONMENT => ', process.env.REACT_APP_ENVIRONMENT);
I build the build with REACT_APP_STAGE and use it in my application as process.env.REACT_APP_STAGE.
"scripts": {
"analyze": "source-map-explorer 'build/static/js/*.js'",
"build-css": "node-sass-chokidar --include-path ./src --include-path ./node_modules src/ -o src/",
"watch-css": "npm run build-css && node-sass-chokidar --include-path ./src --include-path ./node_modules src/ -o src/ --watch --recursive",
"start-js": "react-scripts start",
"start": "REACT_APP_STAGE=local npm-run-all -p watch-css start-js",
"build": "npm run build-css && react-scripts build",
"build-dev": "REACT_APP_STAGE=dev react-scripts build",
"build-prod": "REACT_APP_STAGE=prod react-scripts build",
"build-qa": "REACT_APP_STAGE=qa react-scripts build",
"test": "react-scripts test --env=jsdom",
"eject": "react-scripts eject"
},
Use cross-env in front of NODE_ENV.
npm i -g cross-env
"staging": "concurrently -k \"npm:staging-server\" \"cross-env NODE_ENV='staging' PORT=3003 react-scripts start\"",
Easiest approach is to add it directly in your command:
"scripts": {
"start": "./node_modules/.bin/nodemon server.js",
"start:prod": "NODE_ENV=prod node server.js",
},
I use React.js, Windows in my webpage project and want to apply the settings of webpack.config.prod.js to the project.
So I used command below at Webstorm terminal.
set NODE_ENV=production
.\node_modules\.bin\webpack --config webpack.config.prod.js
yarn build
serve -p 3000 -s build
This below is part of package.json
"scripts": {
"start": "cross-env NODE_PATH=src node scripts/start.js",
"build": "node scripts/build.js",
"test": "node scripts/test.js --env=jsdom",
"development": "cross-env NODE_PATH=src node scripts/start.js"
},
Is it correct command to apply the webpack.config.prod.js to my create-react-app? I don't know how to check that the configuration of webpack.config.prod.js has been applied.
I also wonder if I need webpack-dev-server to apply webpack.config.prod.js.
Any help will be thankful. Thanks for reading.
"build": "node scripts/build.js",
instead of this
use
NODE_ENV=production .\node_modules\.bin\webpack --config webpack.config.prod.js
It will build with production config
Now run yarn run build