unmarshalling 相关问题

我想用Jaxb反序列化xml并使用生成器类下面是我的xml输入嘿 [嘿，你好吗？这里是在JAXB中使用不可变的类和构建器模式的示例。 @XmlRootElement(name = "Root") @XmlType(propOrder = {"head", "message"}) public class Root { @XmlRootElement(name = "Root") @XmlType(name = "RootBuilder") public static class Builder { private String head; private Message.Builder message; @XmlElement(name = "Head") public Builder setName(String head) { this.head = head; return this; } @XmlElement(name = "Message") public Builder setMessage(Message.Builder message) { this.message = message; return this; } public Root build() { return new Root(this.head, this.message.build()); } } private final String head; private final Message message; Root() { // DO NOT REMOVE. This constructor is required by JAXB throw new UnsupportedOperationException(); } Root(String head, Message message) { this.head = head; this.message = message; } @XmlElement(name = "Head") public String getHead() { return this.head; } @XmlElement(name = "Message") public Message getMessage() { return this.message; } @Override public String toString() { return "Root[head=" + this.head + ", message=" + this.message + "]"; } } public class Message { @XmlType(name = "MessageBuilder") public static class Builder { private String code; @XmlAttribute public Builder setCode(String code) { this.code = code; return this; } public Message build() { return new Message(this.code); } } private final String code; Message(String code) { this.code = code; } @XmlAttribute public String getCode() { return this.code; } @Override public String toString() { return "Message[code=" + this.code + "]"; } } Test Root root1 = new Root.Builder() .setName("Hey") .setMessage(new Message.Builder().setCode("MNP")) .build(); System.out.println(root1); StringWriter stringWriter = new StringWriter(); Marshaller marshaller = JAXBContext.newInstance(Root.class).createMarshaller(); marshaller.setProperty(Marshaller.JAXB_FRAGMENT, true); marshaller.marshal(root1, stringWriter); String xml = stringWriter.toString(); System.out.println(xml); Unmarshaller unmarshaller = JAXBContext.newInstance(Root.Builder.class).createUnmarshaller(); Root.Builder rootBuilder = (Root.Builder) unmarshaller.unmarshal(new StringReader(xml)); Root root2 = rootBuilder.build(); System.out.println(root2); 输出 Root[head=Hey, message=Message[code=MNP]] <Root><Head>Hey</Head><Message code="MNP"/></Root> Root[head=Hey, message=Message[code=MNP]]

java jaxb unmarshalling xml-deserialization

回答 1 投票 0

无法从XML中解组非字符串值的字段

我有这样的xml响应

java xml unmarshalling

回答 2 投票 1

如何在Go中读取JSON对象而不对其进行解码（用于读取大数据流）

我正在读取JSON以响应HTTP端点，并希望提取嵌套在其中的对象数组的内容。响应可能很大，因此我尝试使用流式传输...

json go stream decode unmarshalling

回答 1 投票 -2

无法解组golang回应

我一直在尝试通过将JSON响应解组到结构中来提取一些JSON，但是我不知道为什么它不能正确执行。我也尝试过gjson但结果相同。我错过了什么吗？