将JSON映射到简单Java子类
我正在尝试从Android中的默认Json API转移到GSON(或Jackson)。但我一直在尝试将JSONObject转换为Java对象。我读了很多教程,但没有发现任何有用的东西 我有两个类(它们只是为了简单起见): 我试图映射到将JSON映射到简单Java子类,java,android,json,jackson,gson,Java,Android,Json,Jackson,Gson,我正在尝试从Android中的默认Json API转移到GSON(或Jackson)。但我一直在尝试将JSONObject转换为Java对象。我读了很多教程,但没有发现任何有用的东西 我有两个类(它们只是为了简单起见): 我试图映射到Dog类的JSON如下: { "id" : "1", "Name" : "Fluffy" } 我正在使用以下代码: Gson gson = new Gson(); Dog dog = gson.fromJson(jsonObject.toString(
Dog类的JSON如下:
{
"id" : "1",
"Name" : "Fluffy"
}
我正在使用以下代码:
Gson gson = new Gson();
Dog dog = gson.fromJson(jsonObject.toString(), Dog.class);
Name
映射正常,但id
未映射
如果GSON(或Jackson)库更简单的话,我该如何实现这一点呢?对于Jackson,我使用这段代码
private static ObjectMapper configMapper() {
ObjectMapper mapper = new ObjectMapper();
mapper.setVisibilityChecker(mapper.getSerializationConfig().getDefaultVisibilityChecker()
.withFieldVisibility(JsonAutoDetect.Visibility.PUBLIC_ONLY)
.withGetterVisibility(JsonAutoDetect.Visibility.PUBLIC_ONLY)
.withSetterVisibility(JsonAutoDetect.Visibility.PUBLIC_ONLY)
.withCreatorVisibility(JsonAutoDetect.Visibility.PUBLIC_ONLY));
mapper.configure(DeserializationConfig.Feature.FAIL_ON_UNKNOWN_PROPERTIES, false);
return mapper;
}
private Dog readDog(String json) {
Dog ret = null;
if (json != null) {
ObjectMapper mapper = configMapper();
try {
ret = mapper.readValue(json, Dog.class);
} catch (Exception e) {
Log.e("tag", Log.getStackTraceString(e));
return null;
}
}
return ret;
}
希望它也适用于您。您的代码应该可以正常工作。尝试检查jsonObject.toString()返回的内容。是否匹配实际的json。范例
import org.json.simple.JSONObject;
import org.json.simple.parser.JSONParser;
import com.google.gson.Gson;
import com.google.gson.annotations.SerializedName;
class Animal {
private int id;
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
@Override
public String toString() {
return "Animal [id=" + id + "]";
}
}
class Dog extends Animal{
@SerializedName("Name")
private String name;
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
@Override
public String toString() {
return "Dog [name=" + name + ", Id=" + getId() + "]";
}
}
public class GSonParser {
public static void main(String[] args) throws Exception {
String json = "{\"id\" : \"1\", \"Name\" : \"Fluffy\"}";
JSONParser parser = new JSONParser();
JSONObject jsonObject = (JSONObject) parser.parse(json);
Gson gson = new Gson();
Dog dog = gson.fromJson(jsonObject.toString(), Dog.class);
System.out.println(dog); // Prints "Dog [name=Fluffy, Id=1]"
}
}
尝试调用super();在您的Dog构造函数中,确保:jsonObject.toString()
返回的json与您发布的相同,并且属性名称匹配。尝试将字段公开
import org.json.simple.JSONObject;
import org.json.simple.parser.JSONParser;
import com.google.gson.Gson;
import com.google.gson.annotations.SerializedName;
class Animal {
private int id;
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
@Override
public String toString() {
return "Animal [id=" + id + "]";
}
}
class Dog extends Animal{
@SerializedName("Name")
private String name;
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
@Override
public String toString() {
return "Dog [name=" + name + ", Id=" + getId() + "]";
}
}
public class GSonParser {
public static void main(String[] args) throws Exception {
String json = "{\"id\" : \"1\", \"Name\" : \"Fluffy\"}";
JSONParser parser = new JSONParser();
JSONObject jsonObject = (JSONObject) parser.parse(json);
Gson gson = new Gson();
Dog dog = gson.fromJson(jsonObject.toString(), Dog.class);
System.out.println(dog); // Prints "Dog [name=Fluffy, Id=1]"
}
}