gulp：传递依赖任务返回流作为参数 [英] gulp: passing dependent task return stream as parameter

问题描述

我试图创建两个gulp任务，并且我希望第二个任务能够获取第一个输出流并继续使用插件。

I'm trying to create two gulp tasks, and I'd like the second task to take the first one's output stream and keep applying plugins to it.

我可以将第一个任务的返回值传递给第二个任务吗？

Can I pass the first task's return value to the second task?

以下行不通：

// first task to be run
gulp.task('concat', function() {
  // returning a value to signal this is sync
  return
    gulp.src(['./src/js/*.js'])
      .pipe(concat('app.js'))
      .pipe(gulp.dest('./src'));
};

// second task to be run
// adding dependency
gulp.task('minify', ['concat'], function(stream) {
  // trying to get first task's return stream
  // and continue applying more plugins on it
  stream
    .pipe(uglify())
    .pipe(rename({suffix: '.min'}))
    .pipe(gulp.dest('./dest'));
};

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

有没有办法做到这一点？

Is there any way to do this?

推荐答案

您无法将流传递给其他任务。
，但您可以根据条件使用 gulp-if 模块跳过某些管道方法。

you can't pass stream to other task. but you can use gulp-if module to skip some piped method depending on conditions.

var shouldMinify = (0 <= process.argv.indexOf('--uglify'));

gulp.task('script', function() {
  return gulp.src(['./src/js/*.js'])
      .pipe(concat('app.js'))
      .pipe(gulpif(shouldMinify, uglify())
      .pipe(gulpif(shouldMinify, rename({suffix: '.min'}))
      .pipe(gulp.dest('./dest'));
});

执行这样的任务来缩小

gulp script --minify

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