Detect release / debug in gulp using Visual Studio 2015

Question

I\'ve set up an ASP.NET 5 project in Visual Studio and created a gulpfile.js which I use to build my typescript and less files.

For release builds, I want to uglify

Answer 1

Here is another solution that allows you to pass the configuration as a flag to your gulp task, instead of running the configuration as the task itself. This would allow you to use gulp-if to trigger different procedures from within the task.

Note: This answer assumes that gulp isn't installed globally.

1. Install gulp as a dev dependency

   npm install gulp --save-dev
   

2. Register gulp as an npm script

   {
       "name": "vs-build-test",
       ...
       "scripts": {
           "gulp": "gulp"
       }
   }
   

3. Install yargs to handle custom flags

   npm install yargs --save-dev
   

4. Require yargs in gulpfiles.js with a default for the configuration flag

   var argv = require('yargs').default('configuration', 'Debug').argv
   

5. Create a build task in gulpfile.js

   gulp.task('build', function () {
       console.log(`Build Configuration: ${argv.configuration}`);
   });
   

6. Add a pre-build event to run gulp [Project Properties -> Build Events -> Pre-build Event]

   cmd.exe /c npm run gulp build -- --configuration $(ConfigurationName)
   

---

The reason this can be useful is so that you can use the same tasks but conditionally run only portions of the process. Take for example a task that builds javascript:

gulp.task('build:js', function () {
    let isProduction = ${argv.configuration} === 'Release';

    ...

    return gulp.src(source)
            .pipe(gulpif(!isProduction, sourcemaps.init()))
            .pipe(concat(outputFile, { newLine: ';' }))
            .pipe(gulpif(isProduction, closure(closureFlags)))
            .pipe(gulpif(!isProduction, sourcemaps.write()))
            .pipe(gulp.dest(basePath));
});