在消息响应中获取多个反向斜杠作为发送到Amazon SNS的嵌套信息的一部分 - 我错过了什么吗?

时间:2018-01-12 00:16:51

标签: json go notifications amazon-sns

我正在获取以下写的golang客户端脚本的以下消息响应,以便将信息发送到Amazon SNS:

消息回复:

{\"recipient\":\"test20@test.com\",\"template\":\"welcome_email_v1\",\"type\":\"email\",\"source\":\"noreply@sender.co\",\"user\":{\"first_name\":\"\\\"tester\\\"\",\"last_name\":\"\\\"M\\\"\"}}"

作为上述消息响应的一部分,我不太确定作为userfirst_name的嵌套last_name哈希的一部分的多个反斜杠是否正常。如果我缺少某些东西,那么对此有更多经验的人可以抛出一些亮点吗?

package services

import (
  "encoding/json"
  "os"
  "github.com/aws/aws-sdk-go/aws"
  "github.com/aws/aws-sdk-go/aws/session"
  "github.com/aws/aws-sdk-go/service/sns"
)

var NotificationType= "email"

func snsNotificationSender(template, recipient, firstName, lastName string, account_id int) (*sns.PublishOutput, error) {
  sess := session.Must(session.NewSessionWithOptions(session.Options{
      Config: aws.Config{Region: aws.String(os.Getenv("AWS_REGION"))},
      Profile: os.Getenv("AWS_SNS_PROFILE"),
  }))

  svc := sns.New(sess)
  var source string = os.Getenv("EMAIL_SOURCE")

  type UserInfo struct {
    FirstName string `json:"first_name"`
    LastName string `json:"last_name"`
  }

  type MessageBody struct {
    Recipient string `json:"recipient"`
    Template string `json:"template"`
    NotificationType string `json:"type"`
    Source string `json:"source"`
    UserInfo `json:"user"`
  }

  type Message struct {
    MessageBody
  }

  msg := Message{
    MessageBody: MessageBody{
      Template: template,
      Recipient: recipient,
      Source: source,
      NotificationType: NotificationType,
      UserInfo: UserInfo{
        FirstName: firstName,
        LastName: lastName,
      },
    },
  }

  encoded_message, err := json.Marshal(msg)
  message := string(encoded_message)

  msgParams := &sns.PublishInput{
      MessageStructure: aws.String("type: json"),
      Message: aws.String(message),
      TopicArn: aws.String(os.Getenv("AWS_SNS_ARN_Topic")),
  }

  msgResp, err := svc.Publish(msgParams)

  return msgResp, err
}

谢谢。

1 个答案:

答案 0 :(得分:0)

我在我的请求终点遇到了SNS客户端服务的错误。

我必须传递first_name = tester和last_name = M而不是将它们作为first_name =" tester"和last_name =" M"。