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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 and concat my javascripts, and for debug I want to include my typescript- and maps in my output folder.

Is there a way to 'tell' gulp the current configuration? I've seen some mention of setting the NODE_ENV environment variable, but thus far the solutions I've seen arent optimal; they require using the command line before starting the IDE, etc.

The closest solution I've seen is here: http://www.colinsalmcorner.com/post/gulp--workaround-for-handling-vs-solution-configuration

This does, however, mean that I can no longer utilize the Task Runner Explorer which is built-in in the IDE

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I know this has been around for awhile but I recently ran into the same issue and was unhappy with the other solutions and workarounds that I found in answers here on SO. I found a very nice solution that we went with in the comments on the aspnet github page: https://github.com/aspnet/Home/issues/1231#issuecomment-182017985

I hope that it helps someone get to a solution they're happier with faster than I did.

Simply, add the following line to your pre-build event in project config

echo {"config" : "$(ConfigurationName)"} > ../buildConfig.json

With that little beauty sitting around you can read it in your tasks and react accordingly. I use it to prevent minifying files when I'm in debug mode like so:

gulp.task("min:js:bundlesFolder", function ()
    {
        var json = fs.readFileSync("./buildConfig.json", "utf8");
        var host = JSON.parse(json.replace(/^uFEFF/, ''));
        host.isDebug = host.config == "Debug";
        console.log(host.isDebug);
        if (host.isDebug === true)
        {
            return;
        }
        return gulp.src([paths.jsbundleFolder + "/**/*.js", "!" + paths.jsbundleFolder + "/**/*.min.js"])
        .pipe(uglify())
        .pipe(gulp.dest(paths.jsbundleFolder));
    });

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