如何一个接一个地依次运行 Gulp 任务

问题描述

在这样的片段中:

gulp.task "coffee", ->
    gulp.src("src/server/**/*.coffee")
        .pipe(coffee {bare: true}).on("error",gutil.log)
        .pipe(gulp.dest "bin")

gulp.task "clean",->
    gulp.src("bin", {read:false})
        .pipe clean
            force:true

gulp.task 'develop',['clean','coffee'], ->
    console.log "run something else"

在 develop 任务中，我想运行 clean 并在完成后运行 coffee ，完成后运行其他内容.但我无法弄清楚.这块不行.请指教.

In develop task I want to run clean and after it's done, run coffee and when that's done, run something else. But I can't figure that out. This piece doesn't work. Please advise.

推荐答案

它还没有正式发布，但是即将推出的 Gulp 4.0 让您可以轻松地使用 gulp.series 执行同步任务.你可以这样做:

It's not an official release yet, but the coming up Gulp 4.0 lets you easily do synchronous tasks with gulp.series. You can simply do it like this:

gulp.task('develop', gulp.series('clean', 'coffee'))

我找到了一篇很好的博客文章，介绍了如何升级和使用这些简洁的功能:通过示例迁移到 gulp 4

I found a good blog post introducing how to upgrade and make a use of those neat features: migrating to gulp 4 by example

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