使用 Jackson 解析 XML 时出现 MismatchedInputException答案

【问题标题】：MismatchedInputException when Parsing XML using Jackson使用 Jackson 解析 XML 时出现 MismatchedInputException
【发布时间】：2019-09-16 06:00:27
【问题描述】：

谁能帮我找出我在这里做错了什么。我必须解析下面的 XML 字符串

<?xml version="1.0" encoding="UTF-8"?>
<Person>
   <isActive>true</isActive>
   <id>A01222262</id>
   <name>Bob</name>
   <addresses>
      <address>
         <attributeName>NEWYORK_ADDRESS</attributeName>
         <attributeValue>NY 10003</attributeValue>
      </address>
      <address>
         <attributeName>CALIFORNIA_ADDRESS</attributeName>
         <attributeValue>CA 92336</attributeValue>
      </address>
   </addresses>
</Person>

为此，我要映射三个 bean，如下所示

@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(
        name = "",
        propOrder = {"isActive", "id", "name", "addresses"}
)
@XmlRootElement(
        name = "Person"
)
public class Person {
    protected boolean isActive;
    @XmlElement(
            name = "Id"
    )
    protected String id;
    protected String name;
    protected Addresses addresses;
// All Setter and Getter
}

地址类如下所示

    @XmlAccessorType(XmlAccessType.FIELD)
    @XmlType(
            name = "Addresses",
            propOrder = {"address"}
    )
    public class Addresses {
        @XmlElement(
                required = true
        )
        protected List<Address> address;
    //All Setter Getter 
}

地址类如下所示

    @XmlAccessorType(XmlAccessType.FIELD)
    @XmlType(
            name = "",
            propOrder = {"attributeName", "attributeValue"}
    )
    public class Address {
        @XmlElement(
                required = true
        )
        protected String attributeName;
        @XmlElement(
                required = true
        )
        protected String attributeValue;
    //All Setter Getter 
}

这是我在解析时遇到的异常

com.fasterxml.jackson.databind.exc.MismatchedInputException: 不能构造console.interceptor.Address 的实例（尽管至少一个创建者存在）：没有字符串参数构造函数/工厂方法从 [Source: （字符串读取器）；行：1，列：173]（通过参考链： console.interceptor.Person["addresses"]->console.interceptor.Addresses["address"]->java.util.ArrayList[0])

我正在尝试像下面这样解析

XmlMapper xmlMapper=new XmlMapper();
xmlMapper.enable(DeserializationFeature.ACCEPT_EMPTY_STRING_AS_NULL_OBJECT);
Person p=xmlMapper.readValue(xmlString,Person.class);  // xmlSting

我使用的是 jackson-dataformat-xml 版本 2.9.8。

【问题讨论】：

标签： java xml parsing jackson

【解决方案1】：

我尝试了您的课程并能够像下面这样解析它。我也仔细检查了 2.9.8 版本以确认。干杯！！

public class StackOverFlowTest {
    public static void main(String[] args){
    JacksonXmlModule module = new JacksonXmlModule();
    module.setDefaultUseWrapper(false);
    XmlMapper xmlMapper = new XmlMapper(module);
    String xmlString="<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"yes\"?><Person><isActive>true</isActive><id>A01222262</id><name>Bob</name><addresses><address><attributeName>NEWYORK_ADDRESS</attributeName><attributeValue>NY 10003</attributeValue></address><address><attributeName>CALIFORNIA_ADDRESS</attributeName><attributeValue>CA 92336</attributeValue></address></addresses></Person>";
        try {
        Person p=xmlMapper.readValue(xmlString,Person.class);
        System.out.println(p.toString());
    } catch (IOException e) {
        e.printStackTrace();
    }
}
}

【讨论】：