将glob平铺到一个目录 [英] Flatten glob down to one directory

问题描述

在Gulp中，我使用 gulp.src 从目录中选择每个字体文件：

Within Gulp, I am using gulp.src to select every font file from a directory:

gulp.task('copy-fonts', function() {
   gulp.src('components/**/*.{ttf,woff,eof,svg}')
   .pipe(gulp.dest('build/fonts'));
});

但是，我想将所有这些字体文件并列在一个目录中，而不是从组件目录重新创建整个树。

However, I would like to have all of these font files wind up in one directory side-by-side rather than have the entire tree re-created from the components directory.

在Gulp中，Gulp Utils ，npm-glob APIs并没有真正帮助我，尽管我可以轻松地跳过它。

Looking in the Gulp, Gulp Utils, and npm-glob APIs didn't really help me, though I could've easily skipped by it.

最好的方法是什么？ p>

What would the best way to go about this?

推荐答案

我会用gulp-flatten：

I would use gulp-flatten:

var flatten = require('gulp-flatten');
gulp.task('copy-fonts', function() {
  gulp.src('dependencies/**/*.{ttf,woff,eof,svg}')
  .pipe(flatten())
 .pipe(gulp.dest('build/fonts'));
});

至于这是如何在内部完成的，请检查： https://github.com/armed/gulp-flatten/blob/master/index.js

As to how this is done internally, check: https://github.com/armed/gulp-flatten/blob/master/index.js

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