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我可以毫无问题地设置模块别名:

"baseUrl": "./app",
 "paths": {
   "assets/*": ["assets/*"],
   "components/*": ["components/*"],
   ...
 },           

这行得通。但是,我(实际上是我的客户,将来想将项目的某些部分提取为模块)想要创建子目录作为不同路径的别名(假设所有目录作为值都存在,并且我无法更改实际目录结构):

"paths": {
   "assets/*": ["assets/*"],
   "components/*": ["components/*"],
   "@myapp/api/*": ["services/myapp/*"],
   "@myapp/state/*": ["state/*"],
 },       

撇开这是否是一个好的模式/实践不谈(因为我被要求这样做),这在技术上是否可行,而不涉及物理目录结构(并且不使用第 3 方依赖项)?(我使用的是 Typescript 3.4.3 和 Vscode 1.33.1)

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1 回答 1

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"baseUrl": "./",
"paths": {
  "@myapp/environment": [ "./src/environments/environment" ],
  "@myapp/extensions": [ "./src/extensions" ],
  "@myapp/testing": [ "./src/testing" ],
  "@myapp/constants": [ "./src/app/shared/constants" ],
  "@myapp/data": [ "./src/app/data" ],
  "@myapp/shared": [ "./src/app/shared" ],
  "@myapp/core/*": [ "./src/app/core/*" ],
  "@myapp/*": [ "./src/app/*" ]
}

这对我的项目来说就像一个魅力

于 2019-04-29T11:45:14.727 回答