在写代码时,经常会遇到各转类型之间互相转换,比如json转换为Map,jsonArray转List集合,List集合转json,现在整理一个工具类,方便日后查阅。...Map jsonStrToMap(String s) { Map map = new HashMap(); //注意这里JSONObject引入的是net.sf.json net.sf.json.JSONObject..., value); } } return map; } // 将jsonArray字符串转换成List集合 public static List jsonToList...(List list) { JSON json=(JSON) JSON.toJSON(list); return json; } public static void main...转List集合,和List集合转json时需要注意,使用的是阿里的fastJson.jar包,不要引错了,Maven项目对应引入: com.alibaba
String test = “jdkalkjda|||djkdla|||djlak”;
1、json转map let json = {"name":"ES6","day":"2014","feature":"新特性"}; //json 2 map let map...= new Map(); for(let i in json){ map.set(i,json[i]); } console.log(map); 2、map转json... //map 2 json let map = new Map(); map.set("name","ES6"); map.set("day","2014"); map.set...("feature","新特性"); let json = {}; for(let [k,v] of map){ json[k]=v; } console.log(json); </script
java中map和json互转工具类,注意方法示例的格式(这里是最简单的无嵌套的格式),不满足此格式的不合适: package com.yscredit.sz.util; import com.alibaba.fastjson.JSON...map转json * @param map {title=国务院2号文件, attach=根据中华人民共和国} * @return json {"title...转json出错",ex); } } /** * json转map * @param jsonStr {"title":"国务院2号文件","attach...new RuntimeException("json转map出错",ex); } } /** * List>转json...; }catch (Exception ex){ throw new RuntimeException("List>转json
1.jsonObject 转 map 相关jar包: import java.util.HashMap; import java.util.Iterator; import java.util.Map...; import java.util.Map.Entry; import com.alibaba.fastjson.JSON; import com.alibaba.fastjson.JSONObject...; 代码: JSONObject user = resJson.getJSONObject("user"); Map userMap = new HashMap(...对象:" + userMap.toString()); 2.map 转 jsonObject 代码: //map对象 Map data =new HashMap()...; String x =JSONObject.toJSONString(data); System.out.println("json字符串:"+x); 发布者:全栈程序员栈长,转载请注明出处:https
根据list对象中的某个属性转换成map /** * 将对象中的某个属性作为map的key 将对象本身作为map的value构成成一个map * * @param fieldToKey...必须是obj的field 我们把field的getValue作为map的key * @author mountain 2019-01-07 17:21 */ public...static Map listToMap(List listObj, String fieldToKey) { Map map = new...(fieldVal, obj); } catch (Exception e) { logger.error("将对象中的某个属性作为map的key...将对象本身作为map的value构成成一个map出现异常", e); } } return map; } 发布者:全栈程序员栈长,转载请注明出处
”; public final static String USER = “user”; public static HashMap getMessage(String msg) { HashMap map...”, m[0]); map.put(“toName”, m[1]); map.put(“content”, m[2]); map.put(“type”, m[3]); return map; } public...static String sendContent(String type, Object mapContent) { Map userMap = new HashMap(); userMap.put...(MessageUtil.TYPE, type); userMap.put(MessageUtil.DATA, mapContent); // Map转JSON字符串 Gson gson = new Gson...= new HashMap(); //String[] msgString = msg.toString().split(“_”); //map.put(“x”, msgString[0]); //map.put
1.使用Gson类中的toJson()方法 Gson gson = new Gson(); String listToJsonString = gson.toJson(list); 2.使用JSONArray...json=JSONArray.fromobject(list);在调用json.toString()方法转换成字符串 JSONArray jsa = JSONArray.fromObject(list
1.jsonObject 转 map 相关jar包: import java.util.HashMap; import java.util.Iterator; import java.util.Map...; import java.util.Map.Entry; import com.alibaba.fastjson.JSON; import com.alibaba.fastjson.JSONObject...; 代码: JSONObject user = resJson.getJSONObject("user"); Map userMap = new HashMap(...对象:" + userMap.toString()); 2.map 转 jsonObject 代码: //map对象 Map data =new HashMap()...; String x =JSONObject.toJSONString(data); System.out.println("json字符串:"+x); 版权声明:本文内容由互联网用户自发贡献,该文观点仅代表作者本人
// convert json to map package main import ( "fmt" "encoding/json" ) func main() { b...:= []byte(`{"IP": "192.168.11.22", "name": "SKY"}`) m := make(map[string]string) err := json.Unmarshal...fmt.Println("m:", m) for k,v :=range m { fmt.Println(k, ":", v) } } output: m: map
// map to json package main import ( "encoding/json" "fmt" ) func main() { s :=...[]map[string]interface{}{} m1 := map[string]interface{}{ "name": "John", "age": 10} m2...:= map[string]interface{}{ "name": "Alex", "age": 12} s = append(s, m1, m2) s = append(...s, m2) b, err := json.Marshal(s) if err !...= nil { fmt.Println("json.Marshal failed:", err) return } fmt.Println("b:",
/** * Map转json */ function MapTOJson(map) { var...str = '{'; var i = 1; for (var key in map) { if...(i == Object.keys(map).length) { str += '"' + key + '":"'+ map[key] + '"';...} else { str += '"' + key + '":"' + map[key] + '",'; }
list = JSONArray.fromObject(str); System.out.println(list); for (Map map: list) {...// map转entity PlanDTO dto = JSON.parseObject(JSON.toJSONString(map), PlanDTO.class);...System.out.println(dto.getRepaymentMoney()); JSONArray consumptionArray = (JSONArray)map.get...("consumption"); List consumptionList = (List) consumptionArray.toCollection... getConsumption() { return consumption; } public void setConsumption(List<Consumption
alibaba String转json转map pom com.alibaba 1.2.47 调用 HashMap hashMap = JSON.parseObject
如何把JSON对象转为map对象呢? JSON 对象保存在大括号内。就像在JavaScript中, 对象可以保存多个 键/值 对。Map对象保存键/值对,是键/值对的集合。...maps = (Map)JSON.parse(str); System.out.println(“这个是用JSON类来解析JSON字符串!!!”)...)map).getValue()); } //第二种方式 Map mapTypes = JSON.parseObject(str); System.out.println(“这个是用JSON类的parseObject...mapType = JSON.parseObject(str,Map.class); System.out.println(“这个是用JSON类,指定解析类型,来解析JSON字符串!!!”)...接口的一个实现类 */ Map json = (Map) JSONObject.parse(str); System.out.println(“这个是用JSONObject类的parse方法来解析JSON
为啥有三个依赖,当发现大多数的框架都依赖于jackson来处理json转换的时候就自然而然的当做理所当然了。...POJO序列化为json字符串: 准备一个POJO: @JsonIgnoreProperties(ignoreUnknown = true) class User implements Serializable...mapper.readValue(expected, arrayType); Assert.assertEquals("Ryan", users[0].getName()); jsonArray转换成List...= mapper.getTypeFactory().constructCollectionType(ArrayList.class, User.class); //the sieze of the list...is dependon the str json length although the json content is not the POJO type maybe List userList
json串 转 list<class> 方法 List转JSONArray和JSONArray转List...强烈推介IDEA2020.2破解激活,IntelliJ IDEA 注册码,2020.2 IDEA 激活码 1.List转JSONArray List list = new ArrayList...(); JSONArray array= JSONArray.parseArray(JSON.toJSONString(list)); 2.JSONArray转List JSONArray array...= new JSONArray(); List list = JSONObject.parseArray(array.toJSONString(), EventColAttr.class...); 3.String转JSONArray String st = "[{name:Tim,age:25,sex:male},{name:Tom,age:28,sex:male},{name:Lily
List转Map的demo1: 返回的map类型:Map @Test public void toMap(){ List...list = List.of(new Content("name", "xiaoming"), new Content("age", "18")); Map...; } List转Map的demo2: 返回的类型Map @Test public void toMap(){ List list = List.of(new Content("name", "xiaoming"), new Content("age", "18")); Map<String,Content...}); } 说明:Function.identity()返回一个输出跟输入一样的Lambda表达式对象 dmeo3:通过分组的方式来得到Map Map<String, List
go 对象json转map // 函 数:Obj2map // 概 要: // 参 数: // obj: 传入Obj // 返回值: // mapObj: map对象 //...err: 错误 func Obj2map(obj interface{}) (mapObj map[string]interface{}, err error) { // 结构体转json b, err...:= json.Marshal(obj) if err !...= nil { return nil, err } var result map[string]interface{} if err := json.Unmarshal(b, &result
字符串格式 JSON.toJSONString(Object object) @Test public void fun1(){ List personArrayList...JSON字符串转JavaBean JSON.parseObject(String text,Class clazz) @Test public void fun2(){ String...JSON字符串数组转List集合 JSON.parseArray(String text, Class clazz) @Test public void fun3(){...Map集合转json JSON.toJSONString(Object object) @Test public void fun4(){ Map...MapJson字符串转Map集合 (Map)JSON.parse(Object object) @Test public void fun5(){ String str="{\
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