英文:
Go: How do I pass a JSON response without unmarshalling it
问题
使用Go语言，我正在尝试从多个端点并发地获取几个JSON响应。我想将每个响应附加到结构体或映射中的字段，并将此结构体/映射作为JSON对象返回（前端后端模式）。因此，我将向Go应用程序发出带有某种标识符的Web请求。它将进一步发出多个Web请求，并将数据编译成一个大对象以作为响应返回。
我正在使用Fiber作为我的框架，但任何通用的Web框架都会类似：
app.Get("/requests/:identifier", func(c *fiber.Ctx) error {
identifier := c.Params("identifier")
timeout := 1600 * time.Millisecond
client := httpclient.NewClient(httpclient.WithHTTPTimeout(timeout))
res, err := client.Get("https://www.example.com/endpoint?id=" + identifier, nil)
if err != nil{
logger.Error("Timout value exceeded")
return c.Status(503).SendString("Socket Timeout")
}
logger.Info("Fetch success: ")
// Heimdall returns the standard *http.Response object
body, err := ioutil.ReadAll(res.Body)
code := 200
response := &main_response{
Service1: body,
}
return c.Status(code).JSON(response)
})
我遇到的问题是，我不需要在Go中解组这些数据，因为我对它没有用处（我只是传递它）。我是否必须解组它，以便我可以将其设置为响应结构体中的字段，就像这样？
type main_response struct {
Service1 []byte `json:"service1"`
Service2 map[string]string `json:"service2"`
Service3 map[string]interface{} `json:"service3"`
}
（我尝试了几种不同的方法来实现这一点。尝试使用字节数组似乎会对响应进行Base64编码）
在返回之前，我将想要将该结构体编组为JSON，因此也许我别无选择，因为我想不到一种告诉Go“只编组主结构体，其他一切都已经是JSON”的方法。在这一点上，我几乎觉得最好构建一个字符串。
英文:
Using Go, I'm attempting to fetch a few JSON responses concurrently from multiple endpoints. I'd like to attach each of these responses to fields in a struct or map and return this struct/map as a JSON object. (Backend for Frontend pattern). So I will make a web request to the Go application with some sort of identifier. It will in turn make several web requests, and compile the data into one large object to return as a response.
I'm using Fiber as my framework but any generic web framework would be similar:
app.Get("/requests/:identifier", func(c *fiber.Ctx) error {
identifier := c.Params("identifier")
timeout := 1600 * time.Millisecond
client := httpclient.NewClient(httpclient.WithHTTPTimeout(timeout))
res, err := client.Get("https://www.example.com/endpoint?id=" + identifier, nil)
if err != nil{
logger.Error("Timout value exceeded")
return c.Status(503).SendString("Socket Timeout")
}
logger.Info("Fetch success: ")
// Heimdall returns the standard *http.Response object
body, err := ioutil.ReadAll(res.Body)
code := 200
response := &main_response{
Service1: body,
}
return c.Status(code).JSON(response)
})
The problem I'm having is, I have no need to unmarshal this data in Go, as I have no use for it (I am simply passing it along). Do I have to unmarshal it just so I can set it as a field in my response struct like this?
type main_response struct {
Service1 []byte `json:"service1"`
Service2 map[string]string `json:"service2"`
Service3 map[string]interface{} `json:"service3"`
}
(I've tried a few different ways to accomplish this. Trying to use a byte array seems to base64 encode the responses)
I will want to marshal that struct to JSON before returning it, so perhaps I have little choice as I can't think of a way to tell Go "only marshal the main struct, everything else is already JSON". It almost feels like I'd be better off building a string at this point.
答案1
得分: 4
使用 json.RawMessage 将包含 JSON 的 []byte 直接复制到响应的 JSON 文档中：
type main_response struct {
Service1 json.RawMessage `json:"service1"`
...
}
response := &main_response{
Service1: body,
}
return c.Status(code).JSON(response)
英文:
Use json.RawMessage to copy a []byte containing JSON directly to the response JSON document:
type main_response struct {
Service1 json.RawMessage `json:"service1"`
...
}
response := &main_response{
Service1: body,
}
return c.Status(code).JSON(response)
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论