Question

现在我有了这段代码

var gulp = require('gulp');
var babel = require('gulp-babel');

var babelify = require('babelify');
var browserify = require('browserify');
var source = require('vinyl-source-stream');

gulp.task('watch', function() {
  gulp.watch('./scripts/es6/*.es6', ['modules'])
});

gulp.task('modules', function() {
  browserify({
    entries: './scripts/es6/popup.es6',
    debug: true
  })
    .transform(babelify)
    .bundle()
    .pipe(source('popup.js'))
    .pipe(gulp.dest('./scripts'));
});

现在它只是在更改时编译popup.es6，但我希望它在./scripts/es6/目录中编译任何内容。我知道这是可能的，但你是怎么做到的？

Answer 1

我明白了。我刚刚删除了browserfy并将其简化为此

var gulp = require('gulp');
var sourcemaps = require('gulp-sourcemaps');
var babel = require('gulp-babel');
var gutil = require('gulp-util');

var path = require('path');

gulp.task('babel', function() {
  return gulp.src('./scripts/es6/*.es6')
        .pipe(babel())
        .pipe(gulp.dest('./scripts'));
});

gulp.task('watch', function() {
  gulp.watch('./scripts/es6/*.es6', ['babel']);
});

gulp.task('default', ['watch']);

make gulp watch并将整个目录编译为es6

1 个答案: