java - 如何忽略 JSON 属性的底层结构并将其存储为字符串？

Question

我有一个像这样的 JSON 文件:

{
    "Resources": {
        "HelloWorldFunction": {
            "Type": "AWS::Serverless::Function",
            "Properties": {
                "Handler": "index.handler",
                "Runtime": "nodejs8.10",
                "Events": {
                    "HelloWorldApi": {
                        "Type": "Api",
                        "Properties": {
                            "Path": "/",
                            "Method": "GET"
                        }
                    }
                },
                "Policies": [
                    {
                        "SNSPublishMessagePolicy": {
                            "TopicName": {
                                "Fn::GetAtt": [
                                    "HelloWorldTopic",
                                    "TopicName"
                                ]
                            }
                        }
                    }
                ],
                "Environment": {
                    "Variables": {
                        "SNS_TOPIC_ARN": {
                            "Ref": "HelloWorldTopic"
                        }
                    }
                },
                "CodeUri": "nothing"
            }
        },
        "HelloWorldTopic": {
            "Type": "AWS::SNS::Topic",
            "Properties": {
                "Subscription": [
                    {
                        "Endpoint": "nothing",
                        "Protocol": "email"
                    }
                ]
            }
        }
    }
}

我正在使用 Jackson YAMLFactory 来解析与此 JSON 等效的 YAML 文件。如何以将“资源”内的所有内容存储在单个字符串中的方式解析它？ (我想将其保留为单独的 YAML/JSON 以供进一步分析)

Answer 1

ObjectMapper mapper = new ObjectMapper();
String resources =  mapper.readTree(new FileReader(path_to_your_json_file).at("/Resources").asText()

或者类似的东西。