java - 获取 MismatchedInputException : Cannot deserialize instance of Object while parsing JSON to result object

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。

我的类(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();
    }

  }

}

在第一种情况下:

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

上述方法将 JSON 包装在列表中，结果如下:

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

但在这种情况下，我无法使用以下行读取值:

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 JSON 转换为结果对象 Config:

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

但不幸的是，我收到了如下异常:

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>

配置类如下:

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

  private Server server;
  private Client client;
}

服务器类

@AllArgsConstructor
@Getter
@NoArgsConstructor
public class Server {

  @JsonProperty("serverport")
  private String serverPort;

}

配置类

@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 对象以创建配置类以供进一步使用。

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);

但毕竟我收到了如下错误:

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:

修改后我的类看起来像:

客户端类

@Getter
@NoArgsConstructor
public class Client {

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

}

服务器类

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

我为我的配置类创建了一个包装器，例如消除 [ 和 ] 的出现。

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

现在我的配置类看起来像:

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

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

但这并没有解决问题，毕竟我收到了异常:

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"])

如果您提出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 进行解码。请阅读有关 JAXB 的信息。另请注意 XmlElement 和 XmlAttribute 的使用方式。

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();
  }

配置类将是这样的

@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;
    }
}

服务器类

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

客户端类

public class Client {

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

    @XmlElement
    public String serverHost;
}