How To Get Task Name Inside Task In Gulp

I'm using gulp-plumber + gulp-notify and want to put task name in gulp-notify as a title. following is the code i wrote, thanks in advance. gulp.task('SOMETASK', function() { r

Solution 1:

gulp.Gulp.prototype.__runTask = gulp.Gulp.prototype._runTask;
gulp.Gulp.prototype._runTask = function(task) {
  this.currentTask = task;
  this.__runTask(task);
}

gulp.task("someTask", function(){
  console.log( this.currentTask.name );
}

Solution 2:

gulp.task('SOMETASK',function() {
    console.log('Task name:', this.seq.slice(-1)[0]) // Task name: SOMETASK
})

Solution 3:

If you want to monkey-patch Gulp, the following will work with Gulp version 3.9.0:

var _gulpStart = gulp.Gulp.prototype.start;

var _runTask = gulp.Gulp.prototype._runTask;

gulp.Gulp.prototype.start = function (taskName) {
    this.currentStartTaskName = taskName;

    _gulpStart.apply(this, arguments);
};

gulp.Gulp.prototype._runTask = function (task) {
    this.currentRunTaskName = task.name;

    _runTask.apply(this, arguments);
};

gulp.task('jscs', function () {
    console.log('this.currentStartTaskName: ' + this.currentStartTaskName);
    console.log('this.currentRunTaskName: ' + this.currentRunTaskName);
});

gulp.task('jshint', function () {
    console.log('this.currentStartTaskName: ' + this.currentStartTaskName);
    console.log('this.currentRunTaskName: ' + this.currentRunTaskName);
});

gulp.task('build', ['jshint', 'jscs']);

Running gulp build will yield the following console output:

c:\project>gulp build
[16:38:54] Using gulpfile c:\project\gulpfile.js
[16:38:54] Starting 'jshint'...
this.currentStartTaskName: build
this.currentRunTaskName: jshint
[16:38:54] Finished 'jshint' after 244 μs
[16:38:54] Starting 'jscs'...
this.currentStartTaskName: build
this.currentRunTaskName: jscs
[16:38:54] Finished 'jscs' after 152 μs
[16:38:54] Starting 'build'...
[16:38:54] Finished 'build' after 3.54 μs

Solution 4:

let aCallerName = (new Error().stack.match(/ at [^(]+ /g)).map(s => s.replace(/(?: at | )/g,''));
console.log( aCallerName ); 

maybe aCallerName[0] or aCallerName[1]....

Solution 5:

Define the task name outside of the callback and reference it where needed.

var taskName = 'SOMETASK';

gulp.task(taskName, function() {
    return gulp.src(sourcePaths)
        .pipe(plumber({errorHandler: notify.onError({
            message: "<%= error.message %>",
            title: taskName
        })}));
});