reason - 编译 Reason 源文件到与源文件相同的目录

Question

我正在编写一个节点应用程序，我想在其中混合 Reason 和原始 JavaScript。

This section在 bucklescript 文档中对其进行了描述

When user has an existing JS project, it makes sense to output the JS file in the same directory as vanilla JS, the schema added a key called in-source so that generate JS file next to ML file.

我假设这是在 bsconfig.json 中文件？但是 key 应该具有什么值？ The schema documentation没有提到这个选项。

我正在运行 Bucklescript 1.9.1 版 - 因此该功能应该可用(自 1.9.0 起可用)。

我如何使用 in-source选项？

我的 bsconfig.json文件如下所示:

{
  "name": "re-server",
  "version": "0.1.0",
  "bsc-flags": ["-bs-super-errors"],
  "in-source": true,  // I tried adding the key here
  "sources": [{
    "dir": "src",
    "in-source": true  // I tried adding the key here
  }
  ],
  "bs-dependencies" : [
    "bs-express"
  ]
}

Answer 1

它应该在“package-specs”部分:

{
  ...
  "package-specs" : [{
    "module": ...,
    "in-source": true
  }]
}

所以你的 bsconfig.json应该看起来像:

{
  "name": "re-server",
  "version": "0.1.0",
  "bsc-flags": ["-bs-super-errors"],
  "package-specs" : [{
    "module": "commonjs",
    "in-source": true
  }],
  "sources": [{
    "dir": "src"
  }
  ],
  "bs-dependencies" : [
    "bs-express"
  ]
}