如何使用 Go 提供 JSON 响应？
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How to serve up a JSON response using Go?
提问by Matthew Harwood
Question:Currently I'm printing out my response in the func Indexlike this fmt.Fprintf(w, string(response))however, how can I send JSON properly in the request so that it maybe consumed by a view?
问题：目前我正在func Index像这样打印我的响应，fmt.Fprintf(w, string(response))但是如何在请求中正确发送 JSON 以便它可能被视图使用？
package main
import (
"fmt"
"github.com/julienschmidt/httprouter"
"net/http"
"log"
"encoding/json"
)
type Payload struct {
Stuff Data
}
type Data struct {
Fruit Fruits
Veggies Vegetables
}
type Fruits map[string]int
type Vegetables map[string]int
func Index(w http.ResponseWriter, r *http.Request, _ httprouter.Params) {
response, err := getJsonResponse();
if err != nil {
panic(err)
}
fmt.Fprintf(w, string(response))
}
func main() {
router := httprouter.New()
router.GET("/", Index)
log.Fatal(http.ListenAndServe(":8080", router))
}
func getJsonResponse()([]byte, error) {
fruits := make(map[string]int)
fruits["Apples"] = 25
fruits["Oranges"] = 10
vegetables := make(map[string]int)
vegetables["Carrats"] = 10
vegetables["Beets"] = 0
d := Data{fruits, vegetables}
p := Payload{d}
return json.MarshalIndent(p, "", " ")
}
回答by dm03514
You can set your content-type header so clients know to expect json
您可以设置内容类型标头，以便客户知道期望 json
w.Header().Set("Content-Type", "application/json")
Another way to marshal a struct to json is to build an encoder using the http.ResponseWriter
将结构编组为 json 的另一种方法是使用 http.ResponseWriter
// get a payload p := Payload{d}
json.NewEncoder(w).Encode(p)
回答by k3mist
Other users commenting that the Content-Typeis plain/textwhen encoding. You have to set the Content-Typefirst w.Header().Set, then the HTTP response code w.WriteHeader.
其他用户评论说Content-Type是plain/text编码时。你必须设置的Content-Type第一w.Header().Set，那么HTTP响应代码w.WriteHeader。
If you call w.WriteHeaderfirst then call w.Header().Setafter you will get plain/text.
如果你w.WriteHeader先打电话w.Header().Set，然后再打电话，你会得到plain/text。
An example handler might look like this;
示例处理程序可能如下所示；
func SomeHandler(w http.ResponseWriter, r *http.Request) {
data := SomeStruct{}
w.Header().Set("Content-Type", "application/json")
w.WriteHeader(http.StatusCreated)
json.NewEncoder(w).Encode(data)
}
回答by poorva
You can do something like this in you getJsonResponsefunction -
你可以在你的getJsonResponse函数中做这样的事情-
jData, err := json.Marshal(Data)
if err != nil {
// handle error
}
w.Header().Set("Content-Type", "application/json")
w.Write(jData)
回答by Aleks Tkachenko
In gobuffalo.io framework I got it to work like this:
在 gobuffalo.io 框架中，我让它像这样工作：
// say we are in some resource Show action
// some code is omitted
user := &models.User{}
if c.Request().Header.Get("Content-type") == "application/json" {
return c.Render(200, r.JSON(user))
} else {
// Make user available inside the html template
c.Set("user", user)
return c.Render(200, r.HTML("users/show.html"))
}
and then when I want to get JSON response for that resource I have to set "Content-type" to "application/json" and it works.
然后当我想获得该资源的 JSON 响应时，我必须将“内容类型”设置为“应用程序/json”并且它可以工作。
I think Rails has more convenient way to handle multiple response types, I didn't see the same in gobuffalo so far.
我认为 Rails 有更方便的方法来处理多种响应类型，到目前为止我在 gobuffalo 中没有看到相同的方法。
回答by Aleks Tkachenko
You may use this package renderer, I have written to solve this kind of problem, it's a wrapper to serve JSON, JSONP, XML, HTML etc.
你可以使用这个包渲染器，我写来解决这种问题，它是一个包装器来提供 JSON、JSONP、XML、HTML 等。
