Java Json到pojo的转换
如何将以下类型的json转换为java对象Java Json到pojo的转换,java,json,spring-boot,Java,Json,Spring Boot,如何将以下类型的json转换为java对象 { "complaint_Map": { "1000067730": "3011351597604397", "1000067730-06": "10582576134561065" } } 如果有人对此有任何想法,请告诉我们是如何做到的。与杰克逊一起 import com.fasterxml.jackson.databind.ObjectMapper; 试一试 在java中,可以使用Jackson库
{
"complaint_Map": {
"1000067730": "3011351597604397",
"1000067730-06": "10582576134561065"
}
}
如果有人对此有任何想法,请告诉我们是如何做到的。与杰克逊一起
import com.fasterxml.jackson.databind.ObjectMapper;
试一试
在java中,可以使用Jackson库将简单的POJO转换为JSON或从JSON转换为POJO 来自维基百科:
public class ReadWriteJackson {
public static void main(String[] args) throws IOException {
ObjectMapper mapper = new ObjectMapper();
String jsonInput =
"{\"id\":0,\"firstName\":\"Robin\",\"lastName\":\"Wilson\"}";
Person q = mapper.readValue(jsonInput, Person.class);
System.out.println("Read and parsed Person from JSON: " + q);
Person p = new Person("Roger", "Rabbit");
System.out.print("Person object " + p + " as JSON = ");
mapper.writeValue(System.out, p);
}
}
Jackson是用于Java的高性能JSON处理器。它的开发人员赞扬了该库的快速、正确、轻量级和符合人体工程学特性的结合
以下是维基百科的一个例子:
public class ReadWriteJackson {
public static void main(String[] args) throws IOException {
ObjectMapper mapper = new ObjectMapper();
String jsonInput =
"{\"id\":0,\"firstName\":\"Robin\",\"lastName\":\"Wilson\"}";
Person q = mapper.readValue(jsonInput, Person.class);
System.out.println("Read and parsed Person from JSON: " + q);
Person p = new Person("Roger", "Rabbit");
System.out.print("Person object " + p + " as JSON = ");
mapper.writeValue(System.out, p);
}
}
您可以使用来自jackson的ObjectMapper来完成 假设json定义了一个java对象, 然后可以通过
import org.codehaus.jackson.map.ObjectMapper;
YourObject mappingClassObject = new YourObject();
ObjectMapper mapper = new ObjectMapper();
try{
mappingClassObject = mapper.readValue(yourJSON, YourObject.class);
}catch(Exception e){
e.printStackTrace();
}
如果您需要使用
Jackson
库的解决方案,请点击这里
自定义类:
@JsonRootName("complaint_Map")
public class Complaint {
private String firstKey;
private String secondKey;
@JsonProperty("1000067730")
public String getFirstKey() {
return firstKey;
}
@JsonProperty("1000067730")
public void setFirstKey(String firstKey) {
this.firstKey = firstKey;
}
@JsonProperty("1000067730-06")
public String getSecondKey() {
return secondKey;
}
@JsonProperty("1000067730-06")
public void setSecondKey(String secondKey) {
this.secondKey = secondKey;
}
@Override
public String toString() {
return "Complaint{" +
"firstKey='" + firstKey + '\'' +
", secondKey='" + secondKey + '\'' +
'}';
}
}
以及测试方法:
String jsonString = "{\"complaint_Map\":{\"1000067730\":\"3011351597604397\",\"1000067730-06\":\"10582576134561065\"}}";
ObjectMapper mapper = new ObjectMapper();
mapper.configure(DeserializationFeature.UNWRAP_ROOT_VALUE, true);
try {
Complaint complaint = mapper.readValue(jsonString, Complaint.class);
System.out.println(complaint);
} catch (Exception e) {
e.printStackTrace();
}
我使用了以下版本(在Maven pom
):
com.fasterxml.jackson.core
杰克逊数据绑定
2.4.1.3
使用jackson或GSON您尝试过什么?问题被标记为spring boot
的原因是什么?因为在我的spring boot项目中,我需要将用户名基础数据移动到策略基础数据,从RIAK KV存储桶移动到另一台服务器的另一个存储桶,这就是为什么这里不需要spring boot标记!!!
String jsonString = "{\"complaint_Map\":{\"1000067730\":\"3011351597604397\",\"1000067730-06\":\"10582576134561065\"}}";
ObjectMapper mapper = new ObjectMapper();
mapper.configure(DeserializationFeature.UNWRAP_ROOT_VALUE, true);
try {
Complaint complaint = mapper.readValue(jsonString, Complaint.class);
System.out.println(complaint);
} catch (Exception e) {
e.printStackTrace();
}
<dependency>
<groupId>com.fasterxml.jackson.core</groupId>
<artifactId>jackson-databind</artifactId>
<version>2.4.1.3</version>
</dependency>