返回Go中的JSON

huangapple
go评论133阅读模式
英文:

Return JSON in Go

问题

如何在golang中获取未编码为字符串的JSON？我正在使用这个项目将Go代码包装在Python中，以便在AWS Lambda函数中执行。

我的问题是，无论我返回json.Marshal(obj)还是string(json.Marshal(obj))，我都会得到Base64编码的JSON或字符串表示（例如"{\"bar\": \"foo\"}"）。如果您正在使用AWS API网关，这不是一个合适的响应，因为它期望纯粹的JSON，就像在Node.js中返回JSON对象一样。

这是我的代码：

  1. package main
  3. import "C"
  5. import (
  6.   "encoding/json"
  7.   "github.com/eawsy/aws-lambda-go-core/service/lambda/runtime"
  8. )
  10. type Response struct {
  11.   StatusCode int               `json:"statusCode"`
  12.   Headers    map[string]string `json:"headers"`
  13.   Body       string            `json:"body"`
  14. }
  16. func Handle(evt json.RawMessage, ctx *runtime.Context) (interface{}, error) {
  17.   res := &Response{
  18.     StatusCode: 1,
  19.     Headers:    map[string]string{"Content-Type": "application/json"},
  20.     Body:       "Hello World",
  21.   }
  22.   content, _ := json.Marshal(res)
  23.   return string(content), nil
  24. }

这是我从AWS得到的结果：

返回Go中的JSON

英文:

How to I get JSON back from golang without it being encoded as a string? I am using this project to wrap go code in python so that I can execute it in an AWS lambda function.

My issue is that whether I return json.Marshal(obj) or string(json.Marshal(obj)) I either get the base64 encoded json or a string representation (e.g. "{\"bar\": \"foo\"}"). This is not an appropiate response if you are using the AWS API gateway as it expects pure json like you would get if you returned a json object in Node.js.

Here's my code:

  1. package main
  3. import "C"
  5. import (
  6.   "encoding/json"
  7.   "github.com/eawsy/aws-lambda-go-core/service/lambda/runtime"
  8. )
  10. type Response struct {
  11.   StatusCode  int       `json:"statusCode"`
  12.   Headers     map[string]string  `json:"headers"`
  13.   Body        string    `json:"body"`
  14. }
  16. func Handle(evt json.RawMessage, ctx *runtime.Context) (interface{}, error) {
  17.   res := &Response{
  18.     StatusCode: 1,
  19.     Headers: map[string]string{"Content-Type": "application/json"},
  20.     Body: "Hello World",
  21.   }
  22.   content, _ := json.Marshal(res)
  23.   return string(content), nil
  24. }

Here's the result I get from AWS:

返回Go中的JSON

答案1

得分: 9

由于AWS Lambda现在正式支持Go，这里是一个修订后的示例。

  1. package main
  3. import (
  4. 	"github.com/aws/aws-lambda-go/lambda"
  5. )
  7. // Response ...
  8. type Response struct {
  9. 	StatusCode int               `json:"statusCode"`
  10. 	Headers    map[string]string `json:"headers"`
  11. 	Body       string            `json:"body"`
  12. }
  14. // Handle ...
  15. func Handle() (Response, error) {
  16. 	return Response{
  17. 			StatusCode: 1,
  18. 			Headers:    map[string]string{"Content-Type": "application/json"},
  19. 			Body:       "Hello World",
  20. 		},
  21. 		nil
  22. }
  24. func main() {
  25. 	lambda.Start(Handle)
  26. }

(下面的示例使用map[string]interface{}数据类型，无需创建type)

  1. package main
  3. import (
  4. 	"github.com/aws/aws-lambda-go/lambda"
  5. )
  7. // Handle ...
  8. func Handle() (map[string]interface{}, error) {
  9. 	return map[string]interface{}{
  10. 			"statusCode": 1,
  11. 			"headers":    map[string]string{"Content-Type": "application/json"},
  12. 			"body":       "Hello World",
  13. 		},
  14. 		nil
  15. }
  17. func main() {
  18. 	lambda.Start(Handle)
  19. }
英文:

Since AWS Lambda now officially supports Go, here is an revised example.

  1. package main
  3. import (
  4. 	"github.com/aws/aws-lambda-go/lambda"
  5. )
  7. // Response ...
  8. type Response struct {
  9. 	StatusCode int               `json:"statusCode"`
  10. 	Headers    map[string]string `json:"headers"`
  11. 	Body       string            `json:"body"`
  12. }
  14. // Handle ...
  15. func Handle() (Response, error) {
  16. 	return Response{
  17. 			StatusCode: 1,
  18. 			Headers:    map[string]string{"Content-Type": "application/json"},
  19. 			Body:       "Hello World",
  20. 		},
  21. 		nil
  22. }
  24. func main() {
  25. 	lambda.Start(Handle)
  26. }

<br>
(Update below with another example)

You can also just use a map[string]interface{} data type directly without having to create a type.

  1. package main
  3. import (
  4. 	&quot;github.com/aws/aws-lambda-go/lambda&quot;
  5. )
  7. // Handle ...
  8. func Handle() (map[string]interface{}, error) {
  9. 	return map[string]interface{}{
  10. 			&quot;statusCode&quot;: 1,
  11. 			&quot;headers&quot;:    map[string]string{&quot;Content-Type&quot;: &quot;application/json&quot;},
  12. 			&quot;body&quot;:       &quot;Hello World&quot;,
  13. 		},
  14. 		nil
  15. }
  17. func main() {
  18. 	lambda.Start(Handle)
  19. }

答案2

得分: 2

看起来interface{}返回类型是问题的根源。将其强制转换为string会返回一个更像JSON的值。拥有interface{}返回类型让我想到其他地方可能期望一个struct或可序列化的返回值。追踪interface{}或者强制转换为string应该可以解决问题。

英文:

https://play.golang.org/p/l2nrhG9lOA

It looks like the interface{} return type is the culprit. Forcing that to a string returns a more JSON-looking value. Having an interface{} return type makes me think something else is expecting a struct or marshal-able return value. Chase the interface{} or force a string and you should be good.

答案3

得分: 0

json.Marshal返回的是一个纯粹的JSON字符串。看起来你正在使用%q来打印它。请参考https://play.golang.org/p/skvAfk-frO。

英文:

json.Marshal does return a pure JSON string. It looks like you're printing it with %q. See https://play.golang.org/p/skvAfk-frO.

huangapple
  • 本文由 发表于 2017年1月28日 05:13:05
  • 转载请务必保留本文链接：https://go.coder-hub.com/41903075.html
  • amazon-web-services
  • go
  • lambda
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