获取 MismatchedInputException：在将 JSON 解析为结果对象时无法反序列化对象的实例

Question

我想将具有配置属性的 XML 文件解析为 JSON，然后将此 JSON 转换为最终结果对象。

我的 class 看起来像：

@SpringBootApplication
public class AdvancedApplication {

  public static void main(String[] args) {

    SpringApplication.run(AdvancedApplication.class, args);

    XmlMapper xmlMapper = new XmlMapper();
    try {
      List XMLEntries = xmlMapper
          .readValue(new ClassPathResource("configuration.xml")
              .getFile(), List.class);

      ObjectMapper mapper = new ObjectMapper();
      String jsonConfig = mapper.writerWithDefaultPrettyPrinter().writeValueAsString(XMLEntries);

      JsonNode parent = new ObjectMapper().readTree(jsonConfig);
      String content = parent.path("serverport").asText();

      System.out.println(content);

      System.out.println(jsonConfig);
    } catch (IOException e) {
      e.printStackTrace();
    }

  }

}

第一种情况：

List XMLEntries = xmlMapper
          .readValue(new ClassPathResource("configuration.xml")
              .getFile(), List.class);

以上方法将 JSON 包装在列表中，结果如下：

[ {
  "serverport" : "9966"
}, {
  "clientport" : "9999",
  "serverHost" : "localhost"
} ]

但在这种情况下，我无法读取以下行的值：

String content = parent.path("serverport").asText();

因为内容是空的。

最后，我决定以这种特殊方式将我的 JSON 转换为结果对象配置：

Config configObject = mapper.readValue(jsonConfig, Config.class);

但不幸的是，我收到如下异常：

com.fasterxml.jackson.databind.exc.MismatchedInputException: Cannot deserialize instance of `com.javase.advanced.config.Config` out of START_ARRAY token
 at [Source: (String)"[ {
  "serverport" : "9966"
}, {
  "clientport" : "9999",
  "serverHost" : "localhost"
} ]"; line: 1, column: 1]

我的 configuration.xml 文件如下所示：

<?xml version="1.0" encoding="UTF-8"?>
<config>
  <server serverport="9966"/>
  <client clientport="9999">
    <serverHost>localhost</serverHost>
  </client>
</config>

配置class如下：

@NoArgsConstructor
@Getter
@AllArgsConstructor
@ToString
public class Config {

  private Server server;
  private Client client;
}

Server.class

@AllArgsConstructor
@Getter
@NoArgsConstructor
public class Server {

  @JsonProperty("serverport")
  private String serverPort;

}

Config.class

@Getter
@NoArgsConstructor
public class Client {

  @JsonProperty("serverHost")
  private String serverHost;

  @JsonProperty("clientport")
  private String clientPort;
}

我想要实现的是将 configuration.xml 文件解析为 JSON 并将其转换为 Config 对象以创建配置 class 以供进一步使用。

编辑：

在 @pvpkiran 建议之后，我尝试了如下方法：

      Config configValue = xmlMapper.readValue(new ClassPathResource("configuration.xml")
          .getFile(), Config.class);

但毕竟我收到了这样的错误：

org.codehaus.jackson.map.exc.UnrecognizedPropertyException: Unrecognized field "serverport" (Class com.javase.advanced.server.Server), not marked as ignorable
 at [Source: C:\Users\mzawadzki\Desktop\advanced\target\classes\configuration.xml; line: 3, column: 30] (through reference chain: com.javase.advanced.config.Config["server"]->com.javase.advanced.server.Server["serverport"])

编辑 2：

修改后我的 classes 看起来像：

Client.class

@Getter
@NoArgsConstructor
public class Client {

  @JacksonXmlProperty(localName = "clientport")
  private String clientPort;
  @JacksonXmlProperty(localName = "serverHost")
  private String serverHost;

}

Server.class

@AllArgsConstructor
@Getter
@NoArgsConstructor
public class Server {
  @JacksonXmlProperty(localName = "serverport")
  private String serverPort;
}

我为我的配置创建了一个包装器 class 想摆脱 [ 和 ] 的发生。

@Getter
@Setter
@AllArgsConstructor
@NoArgsConstructor
@ToString
@JacksonXmlRootElement(localName = "config")
public class ConfigWrapper {
  private Config[] configurations;
}

现在我的配置 class 看起来像：

@NoArgsConstructor
@Getter
@AllArgsConstructor
@ToString
public class Config {

  @JacksonXmlProperty(localName = "serverport")
  private Server server;
  @JacksonXmlProperty(localName = "client")
  private Client client;
}

但同样没有解决问题，毕竟我收到异常：

org.codehaus.jackson.map.exc.UnrecognizedPropertyException: Unrecognized field "server" (Class com.javase.advanced.config.ConfigWrapper), not marked as ignorable
 at [Source: (through reference chain: com.javase.advanced.config.ConfigWrapper["server"])

如果能提供有关如何实现目标并消除此异常的建议，我将不胜感激。

Answer 1

这应该有效。您可以使用 JAXB 进行解组。请阅读有关 JAXB 的内容。还要注意如何使用 XmlElement 和 XmlAttribute。

String xmlString = "<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n" +
                "<config>\n" +
                "  <server serverport=\"9966\"/>\n" +
                "  <client clientport=\"9999\">\n" +
                "    <serverHost>localhost</serverHost>\n" +
                "  </client>\n" +
                "</config>";

JAXBContext jaxbContext;
 try
   {
      jaxbContext = JAXBContext.newInstance(Config.class);
      Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
      Config config = (Config) jaxbUnmarshaller.unmarshal(new StringReader(xmlString));
      System.out.println(config);
   }catch (JAXBException e){
      e.printStackTrace();
  }

Config class 会像这样

@XmlRootElement(name = "config")
@XmlAccessorType(XmlAccessType.PROPERTY)
public class Config {
    public Server server;
    public Client client;

    public Config() {
    }

    public Server getServer() {
        return server;
    }

    public void setServer(Server server) {
        this.server = server;
    }

    public Client getClient() {
        return client;
    }

    public void setClient(Client client) {
        this.client = client;
    }
}

服务器class

public class Server {
    @XmlAttribute(name = "serverport")
    public String serverPort;
}

客户class

public class Client {

    @XmlAttribute(name = "clientport")
    public String clientPort;

    @XmlElement
    public String serverHost;
}

Answer 2

pvpkiran 的解决方案解决了一个问题，但我无法忍受 Jackson 无法将 XML 解析为单个对象的事实。调查带来了理想的效果，最终证明我在我的 pom.XML 中有两个相似的依赖关系，如 jackson-databind 和 jackson-xml-databind。原来他们之间有冲突。注释掉 jackson-xml-databind 后一切正常。

现在我的类看起来像：

Client.class

@Getter
@NoArgsConstructor
@Setter
public class Client {

  @JacksonXmlProperty(localName = "clientport")
  private String clientPort;
  @JacksonXmlProperty(localName = "serverHost")
  private String serverHost;

}

Server.class

@AllArgsConstructor
@Getter
@NoArgsConstructor
@Setter
public class Server {
  @JacksonXmlProperty(localName = "serverport")
  private String serverPort;
}

Config.class

@AllArgsConstructor
@NoArgsConstructor
@Getter
@Setter
@ToString
@JacksonXmlRootElement(localName = "config")
public class Config {

  private Server server;
  private Client client;
}

再次感谢您的承诺。

获取 MismatchedInputException：在将 JSON 解析为结果对象时无法反序列化对象的实例

Getting MismatchedInputException: Cannot deserialize instance of Object while parsing JSON to result object

java

xml

jaxb