Deserialize JSON

I just started a career in programming, the following is a personal understanding, if some wrong, please indicate in the comments section, if there is a more detailed blog can recommend to me.

 

You must first create an entity classes based on JSON, and to implement the Serializable interface, and then create a JacksonUtils Tools

 

 1 import java.io.Serializable;
 2  
 3  public class Opst implements Serializable{
 4     
 5   
 6      private long dataCount;
 7      private long mediaCount;
 8   
 9      public long getDataCount() {
10          return dataCount;
11      }
12  
13      public void setDataCount(long dataCount) {
14          this.dataCount = dataCount;
15      }
16 
17      
18       public  long getmediacount () {
 19           return mediacount;
20       }
 21   
22       public  void setmediacount ( long mediacount) {
 23           this .mediacount = mediacount;
24       }
 25   
26   }

 1 import com.fasterxml.jackson.databind.ObjectMapper;
 2 import com.fasterxml.jackson.databind.ObjectWriter;
 3 
 4 import java.io.File;
 5 import java.io.IOException;
 6 import java.io.Writer;
 7 
 8 public final class JacksonUtils {
 9 
10     private static final ObjectMapper OBJECT_MAPPER = new ObjectMapper();
11 public static <T> T fromJsonString(String json, Class<T> clazz) {
12         if (json == null)
13             return null;
14         try {
15             return OBJECT_MAPPER.readValue(json, clazz);
16         } catch (Exception e) {
17             throw new RuntimeException("Unable to parse Json String.", e);
18         }
19         return null;
20     }
21 }

 

You can then deserialize, the following is the test code

import com.mhl.common.util.JacksonUtils;

1  public static void main(String[] args) {
2          String a="{\"dataCount\":29833,\"mediaCount\":4123}";
3          Opst opst = JacksonUtils.fromJsonString(a, Opst.class);
4          System.out.println(opst.getDataCount());
5          System.out.println(opst.getMediaCount());
6      }

 

Output

29833

4123