Gulp 可以覆盖所有的 src 文件吗? [英] Can Gulp overwrite all src files?

问题描述

假设我想替换一堆文件中的版本号，其中许多文件位于子目录中.我将通过 gulp-replace 管道文件来运行 regex-replace 函数；但我最终会想要覆盖所有原始文件.

Let's say I want to replace the version number in a bunch of files, many of which live in subdirectories. I will pipe the files through gulp-replace to run the regex-replace function; but I will ultimately want to overwrite all the original files.

任务可能如下所示:

gulp.src([
    './bower.json',
    './package.json',
    './docs/content/data.yml',
    /* ...and so on... */
  ])
  .pipe(replace(/* ...replacement... */))
  .pipe(gulp.dest(/* I DONT KNOW */);

那么我该如何结束它，以便每个 src 文件在其原始位置覆盖自己?有什么我可以传递给 gulp.dest() 的东西吗?

So how can I end it so that each src file just overwrites itself, at its original location? Is there something I can pass to gulp.dest() that will do this?

推荐答案

我能想到两个解决方案:

I can think of two solutions:

1. 将 base 的选项添加到您的 gulp.src 中，如下所示:

1. Add an option for base to your gulp.src like so:

gulp.src([...files...], {base: './'}).pipe(...)...

这将告诉 gulp 保留整个相对路径.然后将 './' 传入 gulp.dest() 以覆盖原始文件.(注意:这是未经测试的，你应该确保你有一个备份，以防它不起作用.)

This will tell gulp to preserve the entire relative path. Then pass './' into gulp.dest() to overwrite the original files. (Note: this is untested, you should make sure you have a backup in case it doesn't work.)

使用函数.Gulp 只是 JavaScript，所以你可以这样做:

Use functions. Gulp's just JavaScript, so you can do this:

[...files...].forEach(function(file) {
    var path = require('path');
    gulp.src(file).pipe(rename(...)).pipe(gulp.dest(path.dirname(file)));
}

如果您需要异步运行这些，第一个会容易得多，因为您需要使用 event-stream.merge 之类的东西并将流映射到数组中.它看起来像

If you need to run these asynchronously, the first will be much easier, as you'll need to use something like event-stream.merge and map the streams into an array. It would look like

var es = require('event-stream');

...

var streams = [...files...].map(function(file) {
        // the same function from above, with a return
        return gulp.src(file) ...
    };
return es.merge.apply(es, streams);

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