JSON仅带有选定字段的Java对象 [英] Java object to JSON with only selected fields

问题描述

我有类似的Java类:

I have java class like:

public class Sample{
 int foo=5;
 int bar=6;
}

现在我要生成JSON对象，但没有栏字段:

now I want to generate JSON object but without bar field:

{"foo":5}

实现此目标的最佳方法是什么? 我应该手动编写JSON字符串，还是可以使用某些库或生成器?

What is a best way to accomplish that? Should I compose JSON string manually, or can I use some library, or generator?

推荐答案

您可以使用Jackson来解决您的问题.请按照以下步骤-

You can use the Jackson to solve your problem. Follow the below step -

步骤1 -制作将Java对象转换为Json的方法

Step 1 - Make a method which will convert Java object to Json

public class JsonUtils {
    public static String javaToJson(Object o) {
            String jsonString = null;
            try {
                ObjectMapper objectMapper = new ObjectMapper();
                objectMapper.configure(org.codehaus.jackson.map.DeserializationConfig.Feature.UNWRAP_ROOT_VALUE,true);  
                jsonString = objectMapper.writeValueAsString(o);

            } catch (JsonGenerationException e) {
                logger.error(e);
            } catch (JsonMappingException e) {
                logger.error(e);
            } catch (IOException e) {
                logger.error(e);
            }
            return jsonString;
        }

}

第2步模型类

package com.javamad.model;

import org.codehaus.jackson.annotate.JsonIgnore;



public class Sample{
     int foo=5;
     public int getFoo() {
        return foo;
    }
    public void setFoo(int foo) {
        this.foo = foo;
    }
    @JsonIgnore
    public int getBar() {
        return bar;
    }
    public void setBar(int bar) {
        this.bar = bar;
    }
    int bar=6;
    }

第3步:将Java类转换为json

Step 3 Convert your java class to json

Sample sample = new Sample()
JsonUtils.javaToJson(sample);

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