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.
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