上传文件的REST API(JSON内部)

huangapple go评论101阅读模式
英文:

REST API for uploading file inside JSON

问题

我正在设计一个 REST API,用于上传一个较大的文件(100MB)以及一些信息。所以自然而然地会考虑使用 JSON 编码。

可以像这样设计:

{
   file: 文件内容或 URL?
   name: 字符串
   description: 字符串
}

name 和 description 可以很容易地使用 JSON 处理,但我不确定如何将文件内容添加到其中。

此外,我在考虑是否应该使用 HTTP PUT 方法。这样做正确吗?

顺便提一下,如果有关系的话,这个 API 是使用 Golang 实现的。

英文:

I am designing an REST API to upload a largish (100MB) file together with some information. So it's natural to think of json encoding.

So something like this:

{
   file: content of the file or URL?
   name: string
   description: string
}

The name and description are easy to do with json but I'm not sure how the file content can be added to it.

Also I'm thinking I should use http PUT method. Is this correct?

Incidentally, golang is used to implement this API if it matters.

答案1

得分: 3

对于JSON编码,使用[]byte值来保存文件内容。标准的encoding/json包将[]byte值编码为base64字符串。

以下是实现JSON编码的草图。声明一个表示有效负载的类型:

type Upload struct {
    Name string
    Description string
    Content []byte
}

将文件编码为请求体:

v := Upload{Name: fileName, Description: description, Content: content}
var buf bytes.Buffer
if err := json.NewEncoder(&buf).Encode(v); err != nil {
    // 处理错误
}
req, err := http.NewRequest("PUT", url, &buf)
if err != nil {
   // 处理错误
}
resp, err := http.DefaultClient.Do(req)

在服务器上从请求体解码:

var v Upload
if err := json.NewDecoder(req.Body).Decode(&v); err != nil {
  // 处理错误
}

另一个选项是使用mime/multipart包。与JSON编码相比,多部分编码更高效,因为不需要对文件进行base64或其他文本编码。

英文:

For a JSON encoding, use a []byte value to hold the file contents. The standard encoding/json package encodes []byte values as base64 strings.

Here's a sketch of how to implement the JSON encoding. Declare a type representing the payload:

type Upload struct {
    Name string
    Description string
    Content []byte
}

To encode the file to a request body:

v := Upload{Name: fileName, Description: description, Content: content}
var buf bytes.Buffer
if err := json.NewEncoder(&buf).Encode(v); err != nil {
    // handle error
}
req, err := http.NewRequest("PUT", url, &buf)
if err != nil {
   // handle error
}
resp, err := http.DefaultClient.Do(req)

To decode the from a request body on the server:

var v Upload
if err := json.NewDecoder(req.Body).Decode(&v); err != nil {
  // handle error
}

Another option is to use the mime/multipart package. The multipart encoding will be more efficient than JSON encoding because no base64 or other text encoding of the file is required for multipart.

答案2

得分: 0

对我来说,最明确的方法是以某种方式对文件字节进行编码。base64似乎是一个不错的选择,并且golang内置了对它的支持,使用"encoding/base64"包即可。

英文:

To me, the most clear-cut way to do it would be to encode the file bytes somehow. base64 seems like a good choice, and golang has built-in support for it with "encoding/base64".

huangapple
  • 本文由 发表于 2014年9月16日 11:54:48
  • 转载请务必保留本文链接:https://go.coder-hub.com/25860391.html
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定