将json串如下:
{
"1":[{"id":001,"age":20,"type":2,"code":"101","password":123456}, {"id":002,"age":22,"type":1,"code":"102","password":123456}],
"2":[{"id":003,"age":23,"type":1,"code":"103","password":123456}]
}
转换为:Map<String, List<User>> 类型

1.  思路就是， json 字符串其实总得来说就是一个 key-value 的形式，应该是满足了最终想要的 Map<String, List<Unit>> 的结构了，所以直接一个 Collectors.toMap 就搞定啦

Map<String, List<User>> result = JSONObject.parseObject(s)
                                           .entrySet().stream()
                                           .collect(Collectors.toMap(Map.Entry::getKey, entry -> JSONObject.parseArray(String.valueOf(entry.getValue()), User.class)));

2. 自己封装一个将 json 解析为常用数据类型的工具类

public static Map<String, Object> jsonToMap(JSONObject json) throws JSONException {
    Map<String, Object> retMap = new HashMap<String, Object>();
    if(json != JSONObject.NULL) {
        retMap = toMap(json);
    return retMap;
public static Map<String, Object> toMap(JSONObject object) throws JSONException {
    Map<String, Object> map = new HashMap<String, Object>();
    Iterator<String> keysItr = object.keys();
    while(keysItr.hasNext()) {
        String key = keysItr.next();
        Object value = object.get(key);
        if(value instanceof JSONArray) {
            value = toList((JSONArray) value);
        else if(value instanceof JSONObject) {
            value = toMap((JSONObject) value);
        map.put(key, value);
    return map;
public static List<Object> toList(JSONArray array) throws JSONException {
    List<Object> list = new ArrayList<Object>();
    for(int i = 0; i < array.length(); i++) {
        Object value = array.get(i);
        if(value instanceof JSONArray) {
            value = toList((JSONArray) value);
        else if(value instanceof JSONObject) {
            value = toMap((JSONObject) value);
        list.add(value);
    return list;
                    将json串如下:{"1":[{"id":001,"age":20,"type":2,"code":"101","password":123456}, {"id":002,"age":22,"type":1,"code":"102","password":123456}],"2":[{"id":003,"age":23,"type":1,"code":"103","password":123...
listData.add(map);
Gson gson = new Gson();
ObjectMapper objectMapper = new ObjectMapper();
List> newMap=objectMapper.r
	 * 传入参数JSON格式校验与转换
	 * @param data 传入JSON格式实字符串
	 * @return Map<String,Object> 返回JSON转换成Map数据
	public st...
import com.alibaba.fastjson.JSON;
import com.alibaba.fastjson.JSONArray;
import com.alibaba.fastjson.JSONObject;
import java.util.List;
import java.util.Map;
 * Created by zkn on 2016/8/22.
public class JsonToM.
List<Map<String, Object>> classifyLh = new ArrayList<>();
Map<String,String> paramInfo = new HashMap<>();
paramInfo.put("batchParam", JSON.toJSONString(classifyLh));//转换
JSON 转L
public static List<Map<String, Object>> toListMap(String json){
        List<Object> list =JSON.parseArray(json);
        List< Map<String,Object>> listw = new ArrayLi..
				
JsonObject和JsonObject转化
1.可以直接将JsonObject对象赋给JsonObject对象
public static void main(String[] args)
	String propertyPairs = "{\"id\":1,\"name\":\"nihh\"}";
	JSONObject property = JSONObject.fromObject(propertyPairs);
	Map<String, Object> propetyMa
				
将list字符串转ArrayList<Map<String,Object>>（需json字符串转map方法）
public static ArrayList<Map<String,Object>> strislist(Object str){
        StringBuffer strb=new StringBuffer((String) str);
        List<String> list=new ArrayList<&gt
利用ObjectMapper类，可将json字符串转成Lis&lt;Map&lt;String,Object&gt;&gt;。
public static void main(String[] args) throws JsonParseException, JsonMappingException, IOException {
String json = "{ \"1\":"...
				
// Json数组形式的字符串转为Map数组
String strArr = "[{"name": "xx", "age": "16"},{"name": "yy", "age": "18"}]";
    //第一种方式
    List<Map<String,String>> listObjectFir = (List<Map<String,String...
* 将json格式的字符串解析成Map对象 
* json格式：{"name":"admin","retries":"3fff","testname"
* :"ddd","testretries":"fffffffff"}
private HashMap toHashMap(JSONObject jsonObject) {
HashMap data = new HashM
				
前提：使用阿里巴巴的JSON包
String json = "";　　//JSON字符串
Map map = JSON.parseObject(json, Map.class);
//遍历 Map 使用迭代器
Iterator it = map.keySet().iterator();
while(it.hashNext()){
String values = map.get(...
				
Jackson是一个Java库，可以将JSON字符串转换为Java对象，例如List<Map>。可以使用ObjectMapper类的readValue()方法将JSON字符串转换为List<Map>。
示例代码:
ObjectMapper mapper = new ObjectMapper();
String jsonString = "{\"name\":\"John\",\"age\":30}";
List<Map<String,Object>> jsonList = mapper.readValue(jsonString, new TypeReference<List<Map<String,Object>>>(){});
jsonString 是你要转换的json字符串
jsonList是经过转换之后的List<Map>
                    weixin_43896630: 
                    public static <K, V> Map<K, List<V>> jsonToMapList(String json, Class<V> object) {
        try {
            TypeFactory typeFactory = OBJECT_MAPPER.getTypeFactory();
            JavaType javaType = typeFactory.constructType(Long.class);
            CollectionType collectionType = typeFactory.constructCollectionType(ArrayList.class, object);
            MapType mapType = typeFactory.constructMapType(HashMap.class, javaType, collectionType);
            return OBJECT_MAPPER.readValue(json, mapType);
        } catch (IOException e) {
            LOGGER.error("JsonUtil.jsonToMapList error:{}", e);
        return Collections.emptyMap();
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