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);

我只用了Gson

HashMap<String, Object> map = new Gson().fromJson(json.toString(), HashMap.class);

简单实用:

/**
 * @param jsonThing can be a <code>JsonObject</code>, a <code>JsonArray</code>,
 *                     a <code>Boolean</code>, a <code>Number</code>,
 *                     a <code>null</code> or a <code>JSONObject.NULL</code>.
 * @return <i>Appropriate Java Object</i>, that may be a <code>Map</code>, a <code>List</code>,
 * a <code>Boolean</code>, a <code>Number</code> or a <code>null</code>.
 */
public static Object jsonThingToAppropriateJavaObject(Object jsonThing) throws JSONException {
    if (jsonThing instanceof JSONArray) {
        final ArrayList<Object> list = new ArrayList<>();

        final JSONArray jsonArray = (JSONArray) jsonThing;
        final int l = jsonArray.length();
        for (int i = 0; i < l; ++i) list.add(jsonThingToAppropriateJavaObject(jsonArray.get(i)));
        return list;
    }

    if (jsonThing instanceof JSONObject) {
        final HashMap<String, Object> map = new HashMap<>();

        final Iterator<String> keysItr = ((JSONObject) jsonThing).keys();
        while (keysItr.hasNext()) {
            final String key = keysItr.next();
            map.put(key, jsonThingToAppropriateJavaObject(((JSONObject) jsonThing).get(key)));
        }
        return map;
    }

    if (JSONObject.NULL.equals(jsonThing)) return null;

    return jsonThing;
}

谢谢@Vikas Gupta。

有一个使用javax的旧答案。json发布在这里，然而它只转换JsonArray和JsonObject，但仍然有JsonString, JsonNumber和JsonValue包装类在输出。如果你想摆脱这些，这是我的解决方案，它会打开所有东西。

除此之外，它还使用Java 8流，并包含在单个方法中。

/**
 * Convert a JsonValue into a “plain” Java structure (using Map and List).
 * 
 * @param value The JsonValue, not <code>null</code>.
 * @return Map, List, String, Number, Boolean, or <code>null</code>.
 */
public static Object toObject(JsonValue value) {
    Objects.requireNonNull(value, "value was null");
    switch (value.getValueType()) {
    case ARRAY:
        return ((JsonArray) value)
                .stream()
                .map(JsonUtils::toObject)
                .collect(Collectors.toList());
    case OBJECT:
        return ((JsonObject) value)
                .entrySet()
                .stream()
                .collect(Collectors.toMap(
                        Entry::getKey,
                        e -> toObject(e.getValue())));
    case STRING:
        return ((JsonString) value).getString();
    case NUMBER:
        return ((JsonNumber) value).numberValue();
    case TRUE:
        return Boolean.TRUE;
    case FALSE:
        return Boolean.FALSE;
    case NULL:
        return null;
    default:
        throw new IllegalArgumentException("Unexpected type: " + value.getValueType());
    }
}

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();