console.log to stdout on gulp events

Javascriptnode.jsGulp

Javascript Problem Overview


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? Many thanks.

Javascript Solutions


Solution 1 - Javascript

(In December 2017, the gulp-util module, which provided logging, was deprecated. The Gulp team recommended that developers replace this functionality with the fancy-log module. This answer has been updated to reflect that.)

fancy-log provides logging and was originally built by the Gulp team.

var log = require('fancy-log');
log('Hello world!');

To add logging, Gulp's API documentation tell us that .src returns:

> Returns a stream of Vinyl files that can be piped to plugins.

Node.js's Stream documentation provides a list of events. Put together, here's an example:

gulp.task('default', function() {
    return gulp.src('main.scss')
        .pipe(sass({ style: 'expanded' }))
        .on('end', function(){ log('Almost there...'); })
        .pipe(minifycss())
        .pipe(gulp.dest('.'))
        .on('end', function(){ log('Done!'); });
});

Note: The end event may be called before the plugin is complete (and has sent all of its own output), because the event is called when "all data has been flushed to the underlying system".

Solution 2 - Javascript

(PLEASE NOTE - In December 2017, the gulp-util module was deprecated.)

To build on the answer by Jacob Budin, I recently tried this and found it useful.

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);
	});
}

Update (April 2021): Re the deprecated module issue. Try replacing util.log from require('gulp-util') with log from require('fancy-log'). The rest of the approach should still work. Disclaimer: I haven't tested this.

Solution 3 - Javascript

Sadly gulp.util was deprecated. Use fancy-log instead: https://www.npmjs.com/package/fancy-log.

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionRimianView Question on Stackoverflow
Solution 1 - JavascriptJacob BudinView Answer on Stackoverflow
Solution 2 - JavascriptTrevedhekView Answer on Stackoverflow
Solution 3 - JavascriptbsesicView Answer on Stackoverflow