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
(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".
Sadly gulp.util
was deprecated.
Use fancy-log
instead:
https://www.npmjs.com/package/fancy-log.
(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);
});
}