可以咕嘟咕嘟覆盖所有的src的文件吗？ [英] Can Gulp overwrite all src files?

问题描述

比方说，我要替换一堆文件，其中许多住在子目录中的版本号。我将通过管道吞掉替换运行正则表达式替换功能;但我最终想要的覆盖所有原始文件的

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.

该任务可能是这个样子：

The task might look something like this:

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

所以，我怎么能结束它让每个的src 文件只是覆盖本身，在其原来的位置？有什么我可以传递给 gulp.dest（）将做到这一点？

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

我能想到的两个解决方案：

I can think of two solutions:

1. 添加一个选项，基你的 gulp.src 像这样：

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

这将告诉一饮而尽，以preserve整个相对路径。然后，只需通过./到 gulp.dest（）来覆盖原文件。 （注：这是未经测试，你应该确保你有一个备份，以防它不工作）

This will tell gulp to preserve the entire relative path. Then simply 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.)

使用功能。咕嘟咕嘟的只是JavaScript，因此你可以这样做：

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)));
}

如果你需要异步运行这些，首先会容易得多，因为你需要使用像事件stream.merge 并映射到数据流数组。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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