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将List数据形式转化为JSON形式

发布时间:2023-09-06 01:16责任编辑:董明明关键词:暂无标签

Action

public String showAllExperts(){ ??????try { ???????list=rescueExpertsService.findRescueExpertsList(); ???????IJsonGenerator json = SpringContextAssist.getBean(IJsonGenerator.class); //获取转机制 ???????toMapList = json.getJsonString(list);//将List转化为Json
??????} catch (Exception e) {
????????????e.printStackTrace();
??????}
?????return "rescue_experts_mapview";
??}
SpringContextAssist
package com.hfky.crs.busi.service;import org.springframework.beans.BeansException;import org.springframework.context.ApplicationContext;import org.springframework.context.ApplicationContextAware;import org.springframework.stereotype.Service;@Servicepublic class SpringContextAssist implements ApplicationContextAware { ?????????private static ApplicationContext context; ?????@Override ?????public void setApplicationContext( ApplicationContext applicationContext ) throws BeansException { ?????????context = applicationContext; ?????} ?????????public static <T> T getBean( String beanName,Class<T> cls ) { ?????????return cls.cast( context.getBean( beanName ) ); ?????} ?????????public static <T> T getBean(Class<T> type){ ???????return context.getBean(type); ???}} 
IJsonGenerator.class
package com.hfky.crs.busi.service;import java.util.List;import com.google.gson.ExclusionStrategy;public interface IJsonGenerator { ???????public IJsonGenerator applyDateFormat(String pattern); ???????/** ????* 将对象实例转换为json字符串 ????* @param target ?待转换的对象 ????* @return ?json字符串 ????*/ ???public String getJsonString(Object target); ???????/** ????* 获取json字符串 ????* @param target ????* @param es ????* @return ????*/ ???public String getJsonString(Object target,ExclusionStrategy es); ???????/** ????* ?????* @param target ????* @param excluded ????* @return ????*/ ???public String getJsonString(Object target,Class<?> ... excluded); ???????/** ????* 从json字符串获取对象实例 ????* @param <Type> 结果类类型 ????* @param json ??json字符串 ????* @param cls ???类型 ????* @return 目标对象 ????*/ ???public <Type> Type getJsonObject(String json,Class<Type> cls ); ???????/** ????* @param <Type> ????* @param json ????* @param cls ????* @return ????*/ ???public <Type> ?List<Type> getListFromJson(String json,Class<Type> cls); ????}

将List数据形式转化为JSON形式

原文地址:http://www.cnblogs.com/flytogalaxy/p/7649924.html

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