如何在 Java 中将 JSON 解析为 Gson 树模型?
javajsonobject oriented programmingprogramming
Gson 库可用于将 JSON 字符串解析为树模型。我们可以使用 JsonParser 将 JSON 字符串解析为 JsonElement 类型的树模型。JsonElement 的 getAsJsonObject() 方法可用于将元素作为 JsonObject 获取，JsonElement 的 getAsJsonArray() 方法可用于将元素作为 JsonArray 获取。
语法
public JsonObject getAsJsonObject() public JsonArray getAsJsonArray()
示例
import java.util.List;
import com.google.gson.*;
public class JsonTreeModelTest {
public static void main(String args[]){
String jsonStr = "{\"name\":\"Adithya\",\"age\":20,\"year of passout\":2005,\"subjects\": [\"MATHEMATICS\",\"PHYSICS\",\"CHEMISTRY\"]}";
JsonParser jsonParser = new JsonParser();
JsonElement jsonElement = jsonParser.parse(jsonStr);
if(jsonElement.isJsonObject()) {
JsonObject studentObj = jsonElement.getAsJsonObject();
System.out.println("Student Info:");
System.out.println("Name is: " + studentObj.get("name"));
System.out.println("Age is: " + studentObj.get("age"));
System.out.println("Year of Passout: " + studentObj.get("year of passout"));
JsonArray jsonArray = studentObj.getAsJsonArray("subjects");
System.out.println("Subjects:" + jsonArray);
}
}
}
// Student classclass Student {
private String name;
private int age;
private int passoutYear;
private List subjects;
public Student(String name, int age, int passoutYear, List subjects) {
this.name = name;
this.age = age;
this.passoutYear = passoutYear;
this.subjects = subjects;
}
@Override
public String toString() {
return "Student{" +
"name='" + name + '\'' +
", age='" + age + '\'' +
", year of passout=" + passoutYear +
", subjects=" + subjects +
'}';
}
}
输出
Student Info: Name is: "Adithya" Age is: 20 Year of Passout: 2005 Subjects:["MATHEMATICS","PHYSICS","CHEMISTRY"]
