如何在Java中忽略JSON对象的多个属性？-java教程-PHP中文网

如何在Java中忽略JSON对象的多个属性？

王林

发布： 2023-09-18 22:29:02

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如何在java中忽略json对象的多个属性？

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@JsonIgnorePropertiesJackson 注释可用于指定要忽略的类的属性列表或字段。 @JsonIgnoreProperties annotation 可以放置在类声明之前，而不是放置在单个属性或字段之前以进行忽略。

语法

@Target(value={ANNOTATION_TYPE,TYPE,METHOD,CONSTRUCTOR,FIELD})
@Retention(value=RUNTIME)
public @interface JsonIgnoreProperties

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示例

import java.io.*;
import com.fasterxml.jackson.annotation.*;
import com.fasterxml.jackson.databind.*;
public class JsonIgnorePropertiesTest {
   public static void main(String[] args) throws IOException {
      Customer customer = new Customer("120", "Ravi", "Hyderabad");
      System.out.println(customer);
      ObjectMapper mapper = new ObjectMapper();
      String jsonString = mapper.writeValueAsString(customer);
      System.out.println("JSON: " + jsonString);
      System.out.println("---------");
      jsonString = "{\"id\":\"130\",\"name\":\"Rahul\", \"address\":\"Mumbai\"}";
      System.out.println("JSON: " + jsonString);
      customer = mapper.readValue(jsonString, Customer.class);
      System.out.println(customer);
   }
}
// Customer class
@JsonIgnoreProperties({"id", "address"}) <strong>
</strong>class Customer {
   private String id;
   private String name;
   private String address;
   public Customer() {
   }
   public Customer(String id, String name, String address) {
      this.id = id;
      this.name = name;
      this.address = address;
   }
   public String getId() {
      return id;
   }
   public String getName() {
      return name;
   }
   public String getAddress() {
      return address;
   }
<strong>   </strong>@Override
   public String toString() {
      return "Customer [id=" + id + ", name=" + name + ", address=" + address + "]";
   }
}

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输出

Customer [id=120, name=Ravi, address=Hyderabad]
JSON: {"name":"Ravi"}
---------
JSON: {"id":"130","name":"Rahul", "address":"Mumbai"}
Customer [id=null, name=Rahul, address=null]

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