Gulp：输出到dest文件夹时排除src根文件夹

Question

Given the following gulpfile, I would expect the contents of the dist folder to be the contents of src/js/ . Instead, the contents of dist is src/js/* .

var del = require('del');
var gulp = require('gulp');
var jshint = require('gulp-jshint');
var livereload = require('gulp-livereload');
var notify = require('gulp-notify');
var sourcemaps = require('gulp-sourcemaps');
var transpile  = require('gulp-es6-module-transpiler');

gulp.task('build', function() {
    return gulp.src('src/js/**/*.js')
        .pipe(sourcemaps.init())
        .pipe(transpile({
            formatter: 'bundle'
        }))
        .pipe(sourcemaps.write('./'))
        .pipe(gulp.dest('dist'))
        .pipe(notify({ message: 'Build task complete.' })); });

How do I exclude the src root directory from the dist output directory, such that the result would be dist/js/**/*.js

Answer 1

My guess is that it's to do with relative paths (compared to where you run gulp from). Try changing the dest to a relative path:

gulp.task('build', function() {
return gulp.src('./src/js/**/*.js')
    .pipe(sourcemaps.init())
    .pipe(transpile({
        formatter: 'bundle'
    }))
    .pipe(sourcemaps.write('./'))
    .pipe(gulp.dest('./dist/js/'))
    .pipe(notify({ message: 'Build task complete.' })); });