使用 jackson xml 映射器将 xml 反序列化为 pojo
deserialize xml to pojo using jackson xml mapper
我正在使用 Jackson XML 映射器将 XML 反序列化为 POJO。 XML 看起来像
<person>
<agency>
<phone>111-111-1111</phone>
</agency>
</person>
我的class看起来像
class Person
{
@JacksonXmlProperty(localName="agency", namespace="namespace")
private Agency agency;
//getter and setter
}
class Agency
{
@JacksonXmlElementWrapper(useWrapping = false)
@JacksonXmlProperty(localName="phone", namespace="namespace")
private List<AgencyPhone> phones;
//getter and setter
}
class AgencyPhone
{
private Phone phone;
//getter and setter
}
class Phone
{
private String number;
//getter and setter
}
我想将 phone 号码设置为 Phone class 中的号码。我无法更改 XML 或 class 的结构方式。我收到 Cannot construct instance of resolved.agency.AgencyPhone
错误,我创建了一个 AgencyPhone constructor
class AgencyPhone{
{
private Phone phone;
public AgencyPhone(Phone phone)
{
this.phone = phone;
}
}
但这没有用。那么如何反序列化为嵌套实例。
您可以编写自己的自定义反序列化器来实现此目的。这是让您入门的代码:
import com.fasterxml.jackson.core.JsonParseException;
import com.fasterxml.jackson.core.JsonParser;
import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.databind.DeserializationContext;
import com.fasterxml.jackson.databind.JsonMappingException;
import com.fasterxml.jackson.databind.deser.std.StdDeserializer;
import com.fasterxml.jackson.databind.module.SimpleModule;
import com.fasterxml.jackson.dataformat.xml.XmlMapper;
import java.io.IOException;
public class Test {
public static void main(String[] args) throws JsonParseException, JsonMappingException, IOException {
XmlMapper mapper = new XmlMapper();
final SimpleModule module = new SimpleModule("configModule", com.fasterxml.jackson.core.Version.unknownVersion());
module.addDeserializer(Person.class, new DeSerializer());
mapper.registerModule(module);
// Person readValue = mapper.readValue(<xml source>);
}
}
class DeSerializer extends StdDeserializer<Person> {
protected DeSerializer() {
super(Person.class);
}
@Override
public Person deserialize(JsonParser p, DeserializationContext ctxt) throws IOException, JsonProcessingException {
// use p.getText() and p.nextToken to navigate through the xml and construct Person object
return new Person();
}
}
我正在使用 Jackson XML 映射器将 XML 反序列化为 POJO。 XML 看起来像
<person>
<agency>
<phone>111-111-1111</phone>
</agency>
</person>
我的class看起来像
class Person
{
@JacksonXmlProperty(localName="agency", namespace="namespace")
private Agency agency;
//getter and setter
}
class Agency
{
@JacksonXmlElementWrapper(useWrapping = false)
@JacksonXmlProperty(localName="phone", namespace="namespace")
private List<AgencyPhone> phones;
//getter and setter
}
class AgencyPhone
{
private Phone phone;
//getter and setter
}
class Phone
{
private String number;
//getter and setter
}
我想将 phone 号码设置为 Phone class 中的号码。我无法更改 XML 或 class 的结构方式。我收到 Cannot construct instance of resolved.agency.AgencyPhone
错误,我创建了一个 AgencyPhone constructor
class AgencyPhone{
{
private Phone phone;
public AgencyPhone(Phone phone)
{
this.phone = phone;
}
}
但这没有用。那么如何反序列化为嵌套实例。
您可以编写自己的自定义反序列化器来实现此目的。这是让您入门的代码:
import com.fasterxml.jackson.core.JsonParseException;
import com.fasterxml.jackson.core.JsonParser;
import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.databind.DeserializationContext;
import com.fasterxml.jackson.databind.JsonMappingException;
import com.fasterxml.jackson.databind.deser.std.StdDeserializer;
import com.fasterxml.jackson.databind.module.SimpleModule;
import com.fasterxml.jackson.dataformat.xml.XmlMapper;
import java.io.IOException;
public class Test {
public static void main(String[] args) throws JsonParseException, JsonMappingException, IOException {
XmlMapper mapper = new XmlMapper();
final SimpleModule module = new SimpleModule("configModule", com.fasterxml.jackson.core.Version.unknownVersion());
module.addDeserializer(Person.class, new DeSerializer());
mapper.registerModule(module);
// Person readValue = mapper.readValue(<xml source>);
}
}
class DeSerializer extends StdDeserializer<Person> {
protected DeSerializer() {
super(Person.class);
}
@Override
public Person deserialize(JsonParser p, DeserializationContext ctxt) throws IOException, JsonProcessingException {
// use p.getText() and p.nextToken to navigate through the xml and construct Person object
return new Person();
}
}