Webpack-Require.context-如何在目录中要求除"_test.js"以外的所有.js文件? [英] Webpack - Require.context -How to require all .js files except `_test.js` in a directory?
问题描述
我的目标是创建一个文件
My goal was to create a file that would
- 要求目录中所有未以
_test.js
结尾的JS文件 - 执行一个
module.exports
等于从那些视图文件返回的模块名称的数组.
- Require all of the JS files in a directory that didn't end in
_test.js
- Do a
module.exports
equal to an array of module names returned from those view files.
我以为我拥有了它:
// Automagically crawls through this directory, finds every js file inside any
// subdirectory, removes test files, and requires the resulting list of files,
// registering the exported module names as dependencies to the myApp.demoApp.views module.
var context = require.context('.', true, /\/.*\/.*\.js$/);
var moduleNames = _.chain(context.keys())
.filter(function(key) {
console.log(key, key.indexOf('_test.js') == -1);
return key.indexOf('_test.js') == -1;
})
.map(function(key) {
console.log("KEY", key);
return context(key)
})
.value();
module.exports = angular.module('myApp.demoApp.views', moduleNames).name;
#2正常工作
#1不幸的是我很幼稚.过滤掉模块名称后,这仍然需要所有带有 _test
的文件,因此测试文件最终将存储在我的内置代码中.
#1 Unfortunately I was naive. While the module names are filtered out, this still requires all of the files with _test
so the test files end up in my built code.
我试图通过更新正则表达式来解决此问题,但JS不支持正则表达式负向隐藏,并且我对正则表达式的理解不足以解决此问题.
I tried to fix this by updating the regex but JS doesn't support regex negative-look-behind and I'm not regex savvy enough to do it without that.
推荐答案
好的,我能够使用Slava.K注释中的答案来回答我的问题.这是下面的最终代码.我必须在正则表达式中包含(?!.* index)
,因为此代码本身包含了 index.views.js
.
Ok, I was able to use the answer in Slava.K's comment to answer my question. Here's the final code below. I had to include (?!.*index)
in the regex because this code was including itself index.views.js
.
var context = require.context('.', true, /^(?!.*index).*\/(?!.*test).*\.js$/);
var moduleNames = _.chain(context.keys())
.map(function(key) {
return context(key)
})
.value();
module.exports = angular.module('myApp.demoApp.views', moduleNames).name;
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