javascript - 如何使用 gulp 正确清理项目？

Question

关于 gulp page有以下例子:

gulp.task('clean', function(cb) {
  // You can use multiple globbing patterns as you would with `gulp.src`
  del(['build'], cb);
});

gulp.task('scripts', ['clean'], function() {
  // Minify and copy all JavaScript (except vendor scripts)
  return gulp.src(paths.scripts)
    .pipe(coffee())
    .pipe(uglify())
    .pipe(concat('all.min.js'))
    .pipe(gulp.dest('build/js'));
});

// Copy all static images
gulp.task('images', ['clean'], function() {
 return gulp.src(paths.images)
    // Pass in options to the task
    .pipe(imagemin({optimizationLevel: 5}))
    .pipe(gulp.dest('build/img'));
});

// the task when a file changes
gulp.task('watch', function() {
  gulp.watch(paths.scripts, ['scripts']);
  gulp.watch(paths.images, ['images']);
});

// The default task (called when you run `gulp` from cli)
gulp.task('default', ['watch', 'scripts', 'images']);

这很好用。但是 watch 任务存在一个大问题。如果我更改图像，监视任务会检测到它并运行 images 任务。这也依赖于 (gulp.task('images', **['clean']**, function() {) clean 任务，所以这个也运行。但是我的脚本文件丢失了，因为 scripts 任务没有再次启动并且 clean 任务删除了所有文件。

我怎样才能在第一次启动时运行清理任务并保留依赖项？

Answer 1

您可以让单独的任务由 watch 触发:

gulp.task('clean', function(cb) {
  // You can use multiple globbing patterns as you would with `gulp.src`
  del(['build'], cb);
});

var scripts = function() {
  // Minify and copy all JavaScript (except vendor scripts)
  return gulp.src(paths.scripts)
    .pipe(coffee())
    .pipe(uglify())
    .pipe(concat('all.min.js'))
    .pipe(gulp.dest('build/js'));
};
gulp.task('scripts', ['clean'], scripts);
gulp.task('scripts-watch', scripts);

// Copy all static images
var images = function() {
 return gulp.src(paths.images)
    // Pass in options to the task
    .pipe(imagemin({optimizationLevel: 5}))
    .pipe(gulp.dest('build/img'));
};
gulp.task('images', ['clean'], images);
gulp.task('images-watch', images);

// the task when a file changes
gulp.task('watch', function() {
  gulp.watch(paths.scripts, ['scripts-watch']);
  gulp.watch(paths.images, ['images-watch']);
});

// The default task (called when you run `gulp` from cli)
gulp.task('default', ['watch', 'scripts', 'images']);