如何使用gulp-browserify观看多个文件，但仅处理一个文件？ [英] How do I watch multiple files with gulp-browserify but process only one?

问题描述

我正在尝试连接 gulp-browserify 和 gulp-watch 在每次源文件更改时重建我的包。但是， gulp-browserify 需要一个编译入口点（例如 src / js / app.js ）和获取每个依赖关系本身：

I'm trying to wire up gulp-browserify and gulp-watch to rebuild my bundle each time a source file changes. However, gulp-browserify requires a single entry point for the compilation (e.g. src/js/app.js) and fetches every dependency itself:

gulp.src('src/js/app.js')
    .pipe(browserify())
    .pipe(gulp.dest('dist'))

然而， ， gulp-watch 在每次更改时都无法重建，因为只有入口点文件正在被监视。我真正需要的是可以观看多个文件，然后只处理入口点文件（查找 replaceEverythingWithEntryPointFile ）：

However, with gulp-watch this fails to rebuild on every change because only the entry point file is being watched. What I actually need is a possibility to watch multiple files and then process only the entry point file (look for replaceEverythingWithEntryPointFile):

gulp.src("src/**/*.js")
    .pipe(watch())
    .pipe(replaceEverythingWithEntryPointFile()) // <- This is what I need
    .pipe(browserify())
    .pipe(gulp.dest("dist"));

所以问题是：我该如何指向 gulp-browserify 到入口点文件，并触发重建任何源文件中的更改？如果解决方案包含节流功能，会很好：在启动时，每个源文件都将被设置为收看，因此我们的入口点文件将被传送到 gulp-browserify 尽可能多次，因为有文件，这是不必要的。

So the question is: how can I point gulp-browserify to the entry point file and trigger rebuild on a change in any source file? Would be nice if the solution included throttling: when starting up, every source file is being set up for watching and thus our entry point file would be piped to gulp-browserify as many times as there are files, which is unnecessary.

推荐答案

只需调用文件更改的普通任务，如下所示：

Just call a normal task on file change, like this:

gulp.task("build-js", function() {
    return gulp.src('src/js/app.js')
        .pipe(browserify())
        .pipe(gulp.dest('dist'))
});

gulp.task("watch", function() {
    // calls "build-js" whenever anything changes
    gulp.watch("src/**/*.js", ["build-js"]);
});

如果您想使用 gulp-watch （因为它可以查找新文件），那么你需要做这样的事情：

If you want to use gulp-watch (because it can look for new files), then you need to do something like this:

gulp.task("watch", function() {
    watch({glob: "src/**/*.js"}, function() {
        gulp.start("build-js");
    });
});

使用 gulp-watch 的批处理操作，所以如果你一次修改几个文件，你不会在一行中获得一堆构建。

Using gulp-watch also has the benefit of batching operations, so if you modify several files at once, you won't get a bunch of builds in a row.

这篇关于如何使用gulp-browserify观看多个文件，但仅处理一个文件？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！