gulp：传递依赖任务返回流作为参数

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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