Java - com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException : Unrecognized field "" not marked as ignorable

Question

关于如何将 XML 转换为 Java pojo 的小问题。

我有一个 super 简单但有效的 xml:

<results preview='0'>
    <messages>
        <msg type="TEST">Why this is failing</msg>
    </messages>
</results>

为了将其转换为 Java pojo，我准备了以下代码片段:

public static void main( String[] args ) throws Exception {
      ObjectMapper objectMapper = new XmlMapper();
      String sss =
              "<results preview='0'>\n" +
              "    <messages>\n" +
              "        <msg type=\"TEST\">Why this is failing</msg>\n" +
              "    </messages>\n" +
              "</results>";
      final MyPojo response = objectMapper.readValue(sss, MyPojo.class);
      System.out.println(response);
  }

使用这个 Java pojo:

public class MyPojo {
    private String preview;
    private Messages messages;

//get set

public class Messages {
    private Msg msg;

//get set

public class Msg {
    private String code;
    private String type;

//get set

然而，当我运行时，我得到:

Exception in thread "main" com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException: Unrecognized field "" (class io.monitoring.Msg), not marked as ignorable (2 known properties: "type", "code"])
 at [Source: (StringReader); line: 3, column: 51] (through reference chain: io.monitoring.MyPojo["messages"]->io.monitoring.Messages["msg"]->io.monitoring.Msg[""])

我可以知道如何解决这个问题吗？ 我对解决异常以及获取所有元素感兴趣，希望获得预览 = 0、类型 = TEST，最重要的是实际消息:为什么失败

谢谢

Answer 1

您需要告诉 xmlMapper 哪个字段是属性，哪个字段是值。请使用下面的注释 @JacksonXmlText 用于您的代码，@JacksonXmlProperty(isAttribute = true) 用于 Msg 类中的类型。像下面这样，

public class Msg {
    @JacksonXmlText
    private String code;
    @JacksonXmlProperty(isAttribute = true)
    private String type;

    public String getCode() {
        return code;
    }

    public void setCode(String code) {
        this.code = code;
    }

    public String getType() {
        return type;
    }

    public void setType(String type) {
        this.type = type;
    }
}