You can convert any JSON to map by using Jackson library as below: String json = "{\r\n\"name\" : \"abc\" ,\r\n\"email id \" : [\"abc@gmail.com\",\"def@gmail.com\",\"ghi@gmail.com\"]\r\n}"; ObjectMapper mapper = new ObjectMapper(); Map<String, Object> map = new HashMap<String, Object>(); // convert JSON string to Map map = mapper.readValue(json, new TypeReference<Map<String, Object>>() {}); System.out.println(map); Maven Dependencies for Jackson : <dependency> <groupId>com.fasterxml.jackson.core</groupId> <artifactId>jackson-core</artifactId> <version>2.5.3</version> <scope>compile</scope> </dependency> <dependency> <groupId>com.fasterxml.jackson.core</groupId> <artifactId>jackson-databind</artifactId> <version>2.5.3</version> <scope>compile</scope> </dependency> Hope this will help. Happy coding :)

试试下面的代码:

 Map<String, String> params = new HashMap<String, String>();
                try
                {

                   Iterator<?> keys = jsonObject.keys();

                    while (keys.hasNext())
                    {
                        String key = (String) keys.next();
                        String value = jsonObject.getString(key);
                        params.put(key, value);

                    }


                }
                catch (Exception xx)
                {
                    xx.toString();
                }

以下是移植到JSR 353的Vikas代码:

import java.util.ArrayList;
import java.util.HashMap;
import java.util.Iterator;
import java.util.List;
import java.util.Map;

import javax.json.JsonArray;
import javax.json.JsonException;
import javax.json.JsonObject;

public class JsonUtils {
    public static Map<String, Object> jsonToMap(JsonObject json) {
        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.keySet().iterator();
        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) {
        List<Object> list = new ArrayList<Object>();
        for(int i = 0; i < array.size(); 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;
    }
}

想象一下你有如下的邮件列表。不受任何编程语言的限制，

emailsList = ["abc@gmail.com","def@gmail.com","ghi@gmail.com"]

下面是JAVA代码-用于将json转换为map

JSONObject jsonObj = new JSONObject().put("name","abc").put("email id",emailsList);
Map<String, Object> s = jsonObj.getMap();

你也可以使用Jackson API:

    final String json = "....your json...";
    final ObjectMapper mapper = new ObjectMapper();
    final MapType type = mapper.getTypeFactory().constructMapType(
        Map.class, String.class, Object.class);
    final Map<String, Object> data = mapper.readValue(json, type);

希望这能起作用，试试这个:

import com.fasterxml.jackson.databind.ObjectMapper;
Map<String, Object> response = new ObjectMapper().readValue(str, HashMap.class);

str，你的JSON字符串

就这么简单，如果你想要电子邮件，

String emailIds = response.get("email id").toString();