我想將具有配置屬性的 XML 文件解析為 JSON，然后將此 JSON 轉換為最終結果對象。我的班級看起來像：@SpringBootApplicationpublic 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(); } }}在第一種情況下：List XMLEntries = xmlMapper .readValue(new ClassPathResource("configuration.xml") .getFile(), List.class);上面的方法將 JSON 包裝在列表中，結果如下：[ { "serverport" : "9966"}, { "clientport" : "9999", "serverHost" : "localhost"} ]但在這種情況下，我無法使用以下行讀取值：String content = parent.path("serverport").asText();因為內容是空的。最后，我決定以這種特殊方式將 JSON 轉換為結果對象 Config：Config configObject = mapper.readValue(jsonConfig, Config.class);但不幸的是，我收到了一個異常，例如：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>以及配置類如下：@NoArgsConstructor@Getter@AllArgsConstructor@ToStringpublic class Config { private Server server; private Client client;}我想要實現的是將configuration.xml 文件解析為JSON 并將其轉換為Config 對象以創建配置類以供進一步使用。
2 回答
慕雪6442864
TA貢獻1812條經驗 獲得超5個贊
這應該有效。您可以使用 JAXB 進行解組。請閱讀有關 JAXB 的信息。還要注意如何使用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();
}
配置類將是這樣的
@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;
}
}
服務器類
public class Server {
@XmlAttribute(name = "serverport")
public String serverPort;
}
客戶類
public class Client {
@XmlAttribute(name = "clientport")
public String clientPort;
@XmlElement
public String serverHost;
}
飲歌長嘯
TA貢獻1951條經驗 獲得超3個贊
該解決方案pvpkiran解決了一個問題，但我無法忍受 Jackson 無法將 XML 解析為單個對象的事實。調查帶來了理想的效果，最終發現 我的 pom.XML 中有兩個類似的依賴項，如jackson-databind和。jackson-xml-databind原來他們之間有矛盾。注釋掉后jackson-xml-databind一切正常。
現在我的課程看起來像：
客戶端類
@Getter
@NoArgsConstructor
@Setter
public class Client {
@JacksonXmlProperty(localName = "clientport")
private String clientPort;
@JacksonXmlProperty(localName = "serverHost")
private String serverHost;
}
服務器類
@AllArgsConstructor
@Getter
@NoArgsConstructor
@Setter
public class Server {
@JacksonXmlProperty(localName = "serverport")
private String serverPort;
}
配置類
@AllArgsConstructor
@NoArgsConstructor
@Getter
@Setter
@ToString
@JacksonXmlRootElement(localName = "config")
public class Config {
private Server server;
private Client client;
}
再次感謝您的承諾。
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