Javascript webpack 需要目录中的每个文件
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webpack require every file in directory
提问by yonatanmn
This is my file structure
这是我的文件结构
-main.js
-SomeDir
-fileA.js
-fileB.js
What should I do if I want to load (inside main.js) every file in someDirwithout specifying the file names -
如果我想main.js在someDir不指定文件名的情况下加载（内部）每个文件，我该怎么办-
something like: require(./someDir/*.js)?
类似的东西：require(./someDir/*.js)？
回答by yonatanmn
Solution:
解决方案：
var req = require.context("../someDir", true, /^(.*\.(js$))[^.]*$/igm);
req.keys().forEach(function(key){
req(key);
});
// or just: req.keys().forEach(req)
extra:
额外的：
regex to match jsbut ignore test.js
正则表达式匹配js但忽略test.js
/^(?!.*test.js)((.*\.(js\.*))[^.]*$)/igm)
/^(?!.*test.js)((.*\.(js\.*))[^.]*$)/igm)
回答by Tudor Ilisoi
Yes, it is possible, even recursively in the folder hierarchy
是的，它是可能的，甚至在文件夹层次结构中递归
Here's a sample:
这是一个示例：
var context = require.context('./', true, /\.(html)$/);
var files={};
context.keys().forEach((filename)=>{
console.log(filename);
files[filename] = context(filename);
});
console.log(files); //you have file contents in the 'files' object, with filenames as keys
And a live demo here (open up devtools to see the generated object)
和一个现场演示（打开 devtools 查看生成的对象）
http://www.webpackbin.com/Vy6AwkWrz
http://www.webpackbin.com/Vy6AwkWrz
In your case the first line would be
在你的情况下，第一行是
var context = require.context('./SomeDir', false, /\.js$/);
Reference: https://webpack.github.io/docs/context.html#require-context
参考：https: //webpack.github.io/docs/context.html#require-context
回答by VyvIT
One approach could be to create an index.js file in SomeDir location and in that file:
一种方法是在 SomeDir 位置和该文件中创建一个 index.js 文件：
var req = require.context('./', true, /\.js$/);
req([]);
Then in main.js:
然后在 main.js 中：
require('./SomeDir');
or
或者
require('./SomeDir/index.js');
or even
甚至
import something from "./SomeDir";
