处理XMLFile超出请求

I created a REST API in GoLang (with gorillamux), and in one request from my API, I process an XMLFile. The problem is, this file is big, 5, 6, 10 MB. I can't process in postfile request, because the time for process. Each node of xml is an http request to another API (very time).

The question is.

In GoLang, its possible receive the request, return response (200), and process file out of request?

In GoLang, its possible receive the request, return response (200), and process file out of request?

Of course. You can use a goroutine:

func myHandler(w http.ResponseWriter, r *http.Request) {
    go func() {
        // Process the stuff in a goroutine
    }()
    w.WriteHeader(200) // but send the response immediately
}

Yes. Go provides very convenient tools for it. One of classical approaches to this task is to make a channel which will serve as a task queue. Your API function will write to the channel new tasks and another worker goroutine will read and process them :

type Task struct {...} // some fields to describe you task - may be XML document 
var TaskQueue chan Task

func worker() {
      for task := range TaskQueue {
             // process task 
       }
}

func handler(w http.ResponseWriter, r*http.Request) {
    task := Task{...} // put here some values you need 
    TaskQueue <- task
    w.WriteHeader(200)
}


func main() {
    TaskQueue = make(chan Task, 1)
     go worker()
     http.Handle("/", handler)
     http.ListenAndServe(":8080", nil)
}

This way gives you some flexibility:

• you may run several workers to process tasks in parallel and you can be sure there will be no more workers so can control load to your server
• if all workers are busy new tasks can be put in queue - second parameter of make sets queue length
• if queue gets overloaded you may check it in handler and return an error (not shown here)