CloudFront Spa子Path，无S3托管

发布于 2025-02-08 16:43:53 字数 641 浏览 1 评论 0原文

我在S3中有一个水疗应用程序。 EC2中的API

我的CloudFront Logic如下：

/-Api-Origin
/front/* - S3-Bucket-Origin

Front具有其方式，例如

example.com/front/front/login/login
example.com/front/documents/
example.com/front/usersetting/

我设法通过托管代理来启动该应用程序。但是我想尝试在不启用S3托管的情况下实现此目标。

我尝试了什么：

为前部行为添加了CloudFront功能。

function handler(event) {
 var request = event.request;
 request.uri = request.uri.replace(/^\/[^/]*\//, "/");
 request.uri += '/front/index.html';

return request;

}

但是我得到404

，请告诉我如何正确实施。最好没有lambda（仅云范围功能）

原文

I have a SPA app in S3. Api in EC2

My cloudfront logic is as follows:

/ - api-origin
/front/* - s3-bucket-origin

The front has its ways, for example

example.com/front/User/Login
example.com/front/Documents/
example.com/front/UserSetting/

I managed to launch the application through s3 bucket hosting proxy. But I want to try to implement this without s3 hosting enabled.

What did I try:

Added cloudfront function for front behavior.

function handler(event) {
 var request = event.request;
 request.uri = request.uri.replace(/^\/[^/]*\//, "/");
 request.uri += '/front/index.html';

return request;

}

But I get 404

Please tell me how to implement this correctly. Preferably without Lambda (only cloudfront functions)

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别在捏我脸啦 2025-02-15 16:43:53

为自己找到了解决方案。我不知道这有多正确，但它有效。

function handler(event) {
  var request = event.request;
  
  request.uri = request.uri.replace(/^\/[^/]*\//, "/");
  
  if (!request.uri.includes('.')) {
    request.uri = "/index.html";
  }

  return request;
}

我希望CloudFront功能不会花很多钱

Found a solution for myself. I don't know how correct this is, but it works.

function handler(event) {
  var request = event.request;
  
  request.uri = request.uri.replace(/^\/[^/]*\//, "/");
  
  if (!request.uri.includes('.')) {
    request.uri = "/index.html";
  }

  return request;
}

I hope cloudfront functions don't cost a lot of money

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