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    1. 如何使用 gulp-browserify 观看多个文件但只处理一个

      时间:2023-05-29
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                本文介绍了如何使用 gulp-browserify 观看多个文件但只处理一个文件?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

                问题描述

                限时送ChatGPT账号..

                我正在尝试连接 gulp-browserifygulp-watch 每次源文件更改.但是,gulp-browserify 需要一个单一的编译入口点(例如 src/js/app.js)并自行获取每个依赖项:

                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'))
                

                但是,使用 gulp-watch 无法在每次更改时重建,因为只监视入口点文件.我真正需要的是可以查看多个文件,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:如何将 gulp-browserify 指向入口点文件并在任何源文件的更改时触发重建?如果解决方案包括限制会很好:启动时,每个源文件都被设置为观看,因此我们的入口点文件将通过管道传输到 gulp-browserify 与文件一样多,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

                只要调用一个正常的文件更改任务,像这样:

                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"]);
                });
                

                如果你想使用 gulp-watch(因为它可以查找新文件),那么你需要这样做:

                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 还具有批处理操作的好处,因此如果您一次修改多个文件,您将不会连续获得一堆构建.

                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 观看多个文件但只处理一个文件?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html5模板网!

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