将JSON转换为Java Object，如何使用Jackson解析BadgerFish约定

Question

使用API我收到这样的JSON（现在保存到文件中）：

[{
    "LEI": {
        "$": "549300Q82NZ9NYNMZT63"
    },
    "Entity": {
        "LegalName": {
            "$": "United Nerds in Collaboration of Random Nerdiness AB"
        },
        "LegalAddress": {
            "Line1": {
                "$": "BOX 155"
            },
            "City": {
                "$": "Alingsas"
            },
            "Region": {
                "$": "SE-O"
            },
            "Country": {
                "$": "SE"
            },
            "PostalCode": {
                "$": "44123"
            }
        },
        "HeadquartersAddress": {
            "Line1": {
                "$": "BOX 155"
            },
            "City": {
                "$": "Alingsas"
            },
            "Region": {
                "$": "SE-O"
            },
            "Country": {
                "$": "SE"
            },
            "PostalCode": {
                "$": "44123"
            }
        },
        "BusinessRegisterEntityID": {
            "@register": "SE001",
            "$": "5568557184"
        },
        "LegalJurisdiction": {
            "$": "SE"
        },
        "LegalForm": {
            "$": "PRIVATA AKTIEBOLAG"
        },
        "EntityStatus": {
            "$": "ACTIVE"
        }
    },
    "Registration": {
        "InitialRegistrationDate": {
            "$": "2016-06-23T01:48:45.025Z"
        },
        "LastUpdateDate": {
            "$": "2016-06-23T01:48:44.945Z"
        },
        "RegistrationStatus": {
            "$": "ISSUED"
        },
        "NextRenewalDate": {
            "$": "2017-06-21T06:32:03.821Z"
        },
        "ManagingLOU": {
            "$": "EVK05KS7XY1DEII3R011"
        },
        "ValidationSources": {
            "$": "PARTIALLY_CORROBORATED"
        }
    }
}]

我想从这些中获取Java Object。我已经从提供的xsd文件中创建了Java对象。我正在运行的代码是：

public static void toJava() {
    ObjectMapper mapper = new ObjectMapper();
    try {
        File json = new File("C:\temp\JSON.json");
        LEIRecordType[] type = mapper.readValue(json, LEIRecordType[].class);
    } catch (JsonEOFException ex) {
        ex.printStackTrace();
    } catch (JsonMappingException ex) {
        ex.printStackTrace();
    } catch (IOException ex) {
        ex.printStackTrace();
    }
}

这会创建以下异常：

com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException: Unrecognized field "LEI" (class   org.leiroc.data.schema.leidata._2014.LEIRecordType), not marked as ignorable (5 known properties: "lei", "registration", "entity", "nextVersion", "extension"])
 at [Source: (File); line: 3, column: 14] (through reference chain: java.lang.Object[][0]->org.leiroc.data.schema.leidata._2014.LEIRecordType["LEI"])
at com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException.from(UnrecognizedPropertyException.java:60)
at com.fasterxml.jackson.databind.DeserializationContext.handleUnknownProperty(DeserializationContext.java:822)
at com.fasterxml.jackson.databind.deser.std.StdDeserializer.handleUnknownProperty(StdDeserializer.java:1152)
at com.fasterxml.jackson.databind.deser.BeanDeserializerBase.handleUnknownProperty(BeanDeserializerBase.java:1567)
at com.fasterxml.jackson.databind.deser.BeanDeserializerBase.handleUnknownVanilla(BeanDeserializerBase.java:1545)
at com.fasterxml.jackson.databind.deser.BeanDeserializer.vanillaDeserialize(BeanDeserializer.java:293)
at com.fasterxml.jackson.databind.deser.BeanDeserializer.deserialize(BeanDeserializer.java:151)
at com.fasterxml.jackson.databind.deser.std.ObjectArrayDeserializer.deserialize(ObjectArrayDeserializer.java:195)
at com.fasterxml.jackson.databind.deser.std.ObjectArrayDeserializer.deserialize(ObjectArrayDeserializer.java:21)
at com.fasterxml.jackson.databind.ObjectMapper._readMapAndClose(ObjectMapper.java:4001)
at com.fasterxml.jackson.databind.ObjectMapper.readValue(ObjectMapper.java:2890)
at Test.JSONParser.toJava(JSONParser.java:38)
at Test.JSONParser.main(JSONParser.java:29)

LEIRecordType如下所示：

package org.leiroc.data.schema.leidata._2014;

import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlType;

@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "LEIRecordType", propOrder = {"lei", "entity", "registration", "nextVersion", "extension"})
public class LEIRecordType {

    @XmlElement(name = "LEI", required = true)
    protected String lei;

    @XmlElement(name = "Entity", required = true)
    protected EntityType entity;

    @XmlElement(name = "Registration", required = true)
    protected RegistrationType registration;

    @XmlElement(name = "NextVersion")
    protected LEIRecordNextVersionType nextVersion;

    @XmlElement(name = "Extension")
    protected ExtensionType extension;

    public String getLEI() {
        return this.lei;
    }

    public void setLEI(String paramString) {
        this.lei = paramString;
    }

    public EntityType getEntity() {
        return this.entity;
    }

    public void setEntity(EntityType paramEntityType) {
        this.entity = paramEntityType;
    }

    public RegistrationType getRegistration() {
        return this.registration;
    }

    public void setRegistration(RegistrationType paramRegistrationType) {
        this.registration = paramRegistrationType;
    }

    public LEIRecordNextVersionType getNextVersion() {
        return this.nextVersion;
    }

    public void setNextVersion(LEIRecordNextVersionType paramLEIRecordNextVersionType) {
        this.nextVersion = paramLEIRecordNextVersionType;
    }

    public ExtensionType getExtension() {
        return this.extension;
    }

    public void setExtension(ExtensionType paramExtensionType) {
        this.extension = paramExtensionType;
    }
}

据我所知，问题是jackson正在锁定一个名为LEI的Java对象，其变量名为“$”。但没有。组织帮助服务说：

“$”对象总是重现相应XML元素的简单内容（即不是属性，子节点等）。“$”对象应始终在适用的地方键入JSON字符串。“

但据我所知，这不是JSON标准。

我的问题是：有没有办法让杰克逊解析这个LEI =“549300Q82NZ9NYNMZT63”等而不是和对象LEI的变量“$”？一天中的大部分时间都被困在这上面。

@UPDATE根据客户服务，这种JSON格式显然被称为“BadgerFish惯例”。

Answer 1

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