无法使用 Fetch TYPESCRIPT 发布多部分/表单数据

Question

我正在尝试获取下面的代码，我开始使用 fetch 所以我遇到了一些问题

import fetch, { Headers, RequestInit } from "node-fetch";
import FormData = require("form-data");

const exampleFile = fs.createReadStream(path.join(__dirname, "../lib/dummy.pdf"));
  const myHeaders = new Headers();
  myHeaders.append("Content-Type", "multipart/form-data");

const form = new FormData();
  form.append("file", exampleFile);

const requestOptions: RequestInit = {
    method: "POST",
    headers: myHeaders,
    body: form,
    redirect: "follow"
  };

await fetch(`https://api.mercadolibre.com/messages/attachments?access_token=${accessToken}`, requestOptions)
    .then(response => response)
    .then(result => console.log(result))
    .catch(error => console.log("error", error));

但这会响应此 JSON（它应该只是来自 MercadoLibre 的附件的 ID）：

Response {
  size: 0,
  timeout: 0,
  [Symbol(Body internals)]: {
    body: PassThrough {
      _readableState: [ReadableState],
      readable: true,
      _events: [Object: null prototype],
      _eventsCount: 2,
      _maxListeners: undefined,
      _writableState: [WritableState],
      writable: false,
      allowHalfOpen: true,
      _transformState: [Object],
      [Symbol(kCapture)]: false
    },
    disturbed: false,
    error: null
  },
  [Symbol(Response internals)]: {
    url: 'https://api.mercadolibre.com/messages/attachments?access_token=#######',
    status: 400,
    statusText: 'Bad Request',
    headers: Headers { [Symbol(map)]: [Object: null prototype] },
    counter: 0
  }
}

我的代码有什么问题吗？

Answer 1

更新：我能够通过排除 Content-Type 属性来修复它，从而允许自动获取检测和设置边界和内容类型。

新代码：

const exampleFile = fs.createReadStream(path.join(__dirname, "../lib/dummy.pdf"));

const form = new FormData();
form.append("file", exampleFile);

const requestOptions: RequestInit = {
    method: "POST",
    body: form
  };

await fetch(`https://api.mercadolibre.com/messages/attachments?access_token=${accessToken}`, requestOptions)
    .then(response => response.json())
    .then(result => console.log(result))
    .catch(error => console.log("error", error));