【发布时间】:2021-05-07 07:32:30
【问题描述】:
请教一个关于如何将 XML 转换为 Java pojo 的小问题。
我有一个超级简单但有效的 xml:
<results preview='0'>
<messages>
<msg type="TEST">Why this is failing</msg>
</messages>
</results>
为了把它转成Java pojo,我准备了这个sn-p:
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,当我跑步时,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[""])
请问我可以知道如何解决这个问题吗? 我有兴趣解决异常以及获取所有元素,希望获得 preview = 0,type = TEST,最重要的是实际消息:为什么会失败
谢谢
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标签: java