随笔 - 836  文章 - 1 评论 - 40 阅读 - 102万
< 2025年3月 >
23 24 25 26 27 28 1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25 26 27 28 29
30 31 1 2 3 4 5

 

 

 

1.jsonUtils

复制代码
package com.icil.utils;

import java.util.List;

import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.databind.JavaType;
import com.fasterxml.jackson.databind.ObjectMapper;
/**

  需要的依赖(jar)

   <!-- Jackson Json处理工具包   用于处理bean 到 json json dao  bean   -->
            <dependency>
                <groupId>com.fasterxml.jackson.core</groupId>
                <artifactId>jackson-databind</artifactId>
                <version>2.4.2</version>
            </dependency>
 
 */


/**
 * 自定义响应结构(pojo to json to pojo)
 */
public class JsonUtils {

    // 定义jackson对象
    private static final ObjectMapper MAPPER = new ObjectMapper();

    /**
     * 将对象转换成json字符串。
     * <p>Title: pojoToJson</p>
     * <p>Description: </p>
     * @param data
     * @return
     */
    public static String objectToJson(Object data) {
        try {
            String string = MAPPER.writeValueAsString(data);
            return string;
        } catch (JsonProcessingException e) {
            e.printStackTrace();
        }
        return null;
    }
    
    /**
     * 将json结果集转化为对象
     *
     * @param jsonData json数据
     * @param clazz 对象中的object类型
     * @return
     */
    public static <T> T jsonToPojo(String jsonData, Class<T> beanType) {
        try {
            T t = MAPPER.readValue(jsonData, beanType);
            return t;
        } catch (Exception e) {
            e.printStackTrace();
        }
        return null;
    }
    
    /**
     * 将json数据转换成pojo对象list
     * <p>Title: jsonToList</p>
     * <p>Description: </p>
     * @param jsonData
     * @param beanType
     * @return
     */
    public static <T>List<T> jsonToList(String jsonData, Class<T> beanType) {
        JavaType javaType = MAPPER.getTypeFactory().constructParametricType(List.class, beanType);
        try {
            List<T> list = MAPPER.readValue(jsonData, javaType);
            return list;
        } catch (Exception e) {
            e.printStackTrace();
        }
        
        return null;
    }
    
}
View Code
复制代码

 

复制代码
package com.icil.esolution.utils;
import java.util.List;

import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.databind.JavaType;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.google.gson.Gson;
import com.google.gson.reflect.TypeToken;

/**
 * 

* @ClassName: JsonUtils 

* @Description:json to bean   object to json 

* @author  Sea

* @date 24 Aug 2018 2:12:27 PM 
*
 */
public class JsonUtils {

    // 定义jackson对象
    private static final ObjectMapper MAPPER = new ObjectMapper();

    /**
     * 将对象转换成json字符串。
     * <p>Title: pojoToJson</p>
     * <p>Description: </p>
     * @param data
     * @return
     */
    public static String objectToJson(Object data) {
        try {
            String string = MAPPER.writeValueAsString(data);
            return string;
        } catch (JsonProcessingException e) {
            e.printStackTrace();
        }
        return null;
    }
    
    /**
     * 将json结果集转化为对象
     *
     * @param jsonData json数据
     * @param clazz 对象中的object类型
     * @return
     */
    public static <T> T jsonToPojo(String jsonData, Class<T> beanType) {
        try {
            T t = MAPPER.readValue(jsonData, beanType);
            return t;
        } catch (Exception e) {
            e.printStackTrace();
        }
        return null;
    }
    
    /**
     * 将json数据转换成pojo对象list
     * <p>Title: jsonToList</p>
     * <p>Description: </p>
     * @param jsonData
     * @param beanType
     * @return
     */
    public static <T>List<T> jsonToList(String jsonData, Class<T> beanType) {
        JavaType javaType = MAPPER.getTypeFactory().constructParametricType(List.class, beanType);
        try {
            List<T> list = MAPPER.readValue(jsonData, javaType);
            return list;
        } catch (Exception e) {
            e.printStackTrace();
        }
        
        return null;
    }
    
    
    /**
     * 可以排除多余的字段,如json中包含字段"aa":"aa" 但是pojo中没有.
     * @param json
     * @param beanType
     * @return
     */
    public static <T>List<T> JsonTOList(String json, Class<T> beanType){
        Gson gson = new Gson();
        List<T> jsonToList =gson.fromJson(json, new TypeToken<List<T>>() {}.getType());
      return jsonToList;    
    }
    
    
    
    
    
}
View Code
复制代码

 

 

 

2.Java对象、Json、Xml转换工具Jackson使用 参考https://blog.csdn.net/Rongbo_J/article/details/47986121

 

依赖:

复制代码
<!-- Jackson Json处理工具包 用于处理bean 到 json json dao bean -->
        <!-- https://mvnrepository.com/artifact/com.fasterxml.jackson.core/jackson-databind -->
        <dependency>
            <groupId>com.fasterxml.jackson.core</groupId>
            <artifactId>jackson-databind</artifactId>
            <version>2.9.6</version>
        </dependency>
        
        <!-- https://mvnrepository.com/artifact/com.fasterxml.jackson.core/jackson-core -->
       <!-- 转json 和json转bean用到-->
     <dependency> <groupId>com.fasterxml.jackson.core</groupId> <artifactId>jackson-core</artifactId> <version>2.9.6</version> </dependency>
<!-- https://mvnrepository.com/artifact/com.fasterxml.jackson.dataformat/jackson-dataformat-xml -->
<!-- 转xml 和xml转bean用到-->
<dependency> <groupId>com.fasterxml.jackson.dataformat</groupId> <artifactId>jackson-dataformat-xml</artifactId> <version>2.9.6</version> </dependency>
复制代码

 

 

 

2.1.將Java对象转换为Json字符串:

复制代码
@Test
    public void testGenJson()
    {
        ObjectMapper objMapper = new ObjectMapper();
        Book book = new Book("Think in Java",100);
        try {
            jsonGen = objMapper.getJsonFactory().createJsonGenerator(System.out,JsonEncoding.UTF8);
            jsonGen.writeObject(book);
        } catch (IOException e) { 
            e.printStackTrace();
        } 
    }
View Code
复制代码

2.2將Json字符串转为Java对象:

复制代码
     /*
     * Java对象转xml
     */
    @Test
    public void testGenXml()
    {
        XmlMapper xmlMapper = new XmlMapper();

        Book book = new Book("Think in Java",100);
        try {
            String xmlStr =  xmlMapper.writeValueAsString(book);
            System.out.println(xmlStr);
        } catch (JsonProcessingException e) { 
            e.printStackTrace();
        }
    }
View Code
复制代码
复制代码
 /*
     * Json转Java对象
     */
    @Test
    public void testGenObjByJson()
    {
        ObjectMapper objMapper = new ObjectMapper();
        String json = "{\"name\":\"Think in Java\",\"price\":100}"; 
        try {
            Book book = objMapper.readValue(json, Book.class);
            System.out.println(book);
        } catch (IOException e) { 
            e.printStackTrace();
        }  
    }
View Code
复制代码

 

 

2.3將Java对象转为Xml格式

复制代码
 /*
     * Java对象转xml
     */
    @Test
    public void testGenXml()
    {
        XmlMapper xmlMapper = new XmlMapper();

        Book book = new Book("Think in Java",100);
        try {
            String xmlStr =  xmlMapper.writeValueAsString(book);
            System.out.println(xmlStr);
        } catch (JsonProcessingException e) { 
            e.printStackTrace();
        }
    }
View Code
复制代码

4.將xml格式字符串转为Java对象

复制代码
  /*
     * xml转Java对象
     */
    @Test
    public void testGenObjByXml()
    {
        XmlMapper xmlMapper = new XmlMapper();
        String xmlStr = "<Book><name>Think in Java</name><price>100</price></Book>"; 
        try {
            Book book = xmlMapper.readValue(xmlStr, Book.class);
            System.out.println(book);
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }

    }
View Code
复制代码

 

posted on   lshan  阅读(886)  评论(0编辑  收藏  举报
编辑推荐:
· .NET Core 中如何实现缓存的预热?
· 从 HTTP 原因短语缺失研究 HTTP/2 和 HTTP/3 的设计差异
· AI与.NET技术实操系列:向量存储与相似性搜索在 .NET 中的实现
· 基于Microsoft.Extensions.AI核心库实现RAG应用
· Linux系列:如何用heaptrack跟踪.NET程序的非托管内存泄露
阅读排行:
· TypeScript + Deepseek 打造卜卦网站:技术与玄学的结合
· Manus的开源复刻OpenManus初探
· AI 智能体引爆开源社区「GitHub 热点速览」
· 三行代码完成国际化适配,妙~啊~
· .NET Core 中如何实现缓存的预热?
点击右上角即可分享
微信分享提示