How to console log to stdout on gulp events

0 votes

I want to log to stdout (the config environment) when a gulp task is running or has run.

Something like this:

gulp.task('scripts', function () {
  var enviroment = argv.env || 'development';
  var config = gulp.src('config/' + enviroment + '.json')
      .pipe(ngConstant({name: 'app.config'}));
  var scripts = gulp.src('js/*');

  return es.merge(config, scripts)
    .pipe(concat('app.js'))
    .pipe(gulp.dest('app/dist'))
    .on('success', function() { 
      console.log('Configured environment: ' + environment);
    });
});

I am not sure what event I should be responding to or where to find a list of these. Any pointers?

Oct 12, 2020 in Node-js by kartik
• 37,520 points
6,432 views

1 answer to this question.

0 votes

Hello @kartik,

Use this:

var gulp = require("gulp");
var util = require("gulp-util");
var changed = require("gulp-changed");

gulp.task("copyIfChanged", function() {
    var nSrc=0, nDes=0, dest="build/js";
    gulp.src("app/**/*.js")
    .on("data", function() { nSrc+=1;})
    .pipe(changed(dest)) //filter out src files not newer than dest
    .pipe(gulp.dest(dest))
    .on("data", function() { nDes+=1;})
    .on("finish", function() {
        util.log("Results for app/**/*.js");
        util.log("# src files: ", nSrc);
        util.log("# dest files:", nDes);
    });
}

Hope it helps!!

Thank you!!

answered Oct 12, 2020 by Niroj
• 82,840 points

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