本文介绍了将glob平铺到一个目录的处理方法,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

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:

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

这篇关于将glob平铺到一个目录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!

10-24 21:52