[英]Parse JSON tree to plain class using Jackson or its alternatives
如何解析那個JSON:
{
"foo": {
"bar": {
"baz": "Hello"
},
"qux": "World"
}
}
使用傑克遜或其替代品進入該類:
public class Foo {
private String baz;
private String qux;
public String getBaz() {
return baz;
}
public void setBaz(final String baz) {
this.baz = baz;
}
public String getQux() {
return qux;
}
public void setQux(final String qux) {
this.qux = qux;
}
}
期待像:
@JsonProperty("foo.bar.baz")
private String baz;
@JsonProperty("foo.qux")
private String qux;
注意:我是EclipseLink JAXB(MOXy)的負責人,也是JAXB(JSR-222)專家組的成員。
Jackson可能無法使用此用例,但可以在將MOXy用作JSON綁定提供程序時完成。
富
您可以利用MOXy基於路徑的映射來完成此用例。
import org.eclipse.persistence.oxm.annotations.XmlPath;
public class Foo {
private String baz;
private String qux;
@XmlPath("foo/bar/baz/text()")
public String getBaz() {
return baz;
}
public void setBaz(final String baz) {
this.baz = baz;
}
@XmlPath("foo/qux/text()")
public String getQux() {
return qux;
}
public void setQux(final String qux) {
this.qux = qux;
}
}
演示
JAXB運行時API用於讀/寫JSON。
import java.util.*;
import javax.xml.bind.*;
import javax.xml.transform.stream.StreamSource;
import org.eclipse.persistence.jaxb.JAXBContextProperties;
public class Demo {
public static void main(String[] args) throws Exception {
Map<String, Object> properties = new HashMap<String, Object>(2);
properties.put(JAXBContextProperties.MEDIA_TYPE, "application/json");
properties.put(JAXBContextProperties.JSON_INCLUDE_ROOT, false);
JAXBContext jc = JAXBContext.newInstance(new Class[] {Foo.class}, properties);
Unmarshaller unmarshaller = jc.createUnmarshaller();
StreamSource json = new StreamSource("src/forum15659950/input.json");
Foo foo = unmarshaller.unmarshal(json, Foo.class).getValue();
Marshaller marshaller = jc.createMarshaller();
marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
marshaller.marshal(foo, System.out);
}
}
input.json /輸出
{
"foo" : {
"bar" : {
"baz" : "Hello"
},
"qux" : "World"
}
}
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