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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更多推荐
gulp:传递依赖任务返回流作为参数
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