如何有效地将org.json.JSONObject映射到POJO？

问题描述

之前一定要问过这个问题，但我找不到。

This question must have been asked before, but I couldn't find it.

我正在使用第三方库来检索JSON格式的数据。该库以 org.json.JSONObject 的形式向我提供数据。我想将此 JSONObject 映射到用于更简单的访问/代码。

I'm using a 3rd party library to retrieve data in JSON format. The library offers the data to me as a org.json.JSONObject. I want to map this JSONObject to a POJO (Plain Old Java Object) for simpler access/code.

对于映射，我目前使用 ObjectMapper 来自这样的Jackson库：

For mapping, I currently use the ObjectMapper from the Jackson library in this way:

JSONObject jsonObject = //...
ObjectMapper mapper = new ObjectMapper();
MyPojoClass myPojo = mapper.readValue(jsonObject.toString(), MyPojoClass.class);

根据我的理解，上面的代码可以显着优化，因为目前<$ c中的数据已经解析的$ c> JSONObject 再次被送入带有 JSONObject.toString（）方法的序列化 - 反序列化链，然后到 ObjectMapper 。

To my understanding, the above code can be optimized significantly, because currently the data in the JSONObject, which is already parsed, is again fed into a serialization-deserialization chain with the JSONObject.toString() method and then to the ObjectMapper.

我想避免这两次转换（ toString（）和解析）。有没有办法使用 JSONObject 将其数据直接映射到POJO？

I want to avoid these two conversions (toString() and parsing). Is there a way to use the JSONObject to map its data directly to a POJO?

推荐答案

由于您有一些JSON数据的抽象表示（ object）你计划使用杰克逊库 - 它有自己的JSON数据抽象表示（） - 然后从一个表示转换到另一个表示将使您免于parse-serialize-parse过程。因此，不使用接受 String 的 readValue 方法，而是使用接受 JsonParser ：

Since you have an abstract representation of some JSON data (an org.json.JSONObject object) and you're planning to use the Jackson library - that has its own abstract representation of JSON data (com.fasterxml.jackson.databind.JsonNode) - then a conversion from one representation to the other would save you from the parse-serialize-parse process. So, instead of using the readValue method that accepts a String, you'd use this version that accepts a JsonParser:

JSONObject jsonObject = //...
JsonNode jsonNode = convertJsonFormat(jsonObject);
ObjectMapper mapper = new ObjectMapper();
MyPojoClass myPojo = mapper.readValue(new TreeTraversingParser(jsonNode), MyPojoClass.class);

JSON是一种非常简单的格式，因此创建 convertJsonFormat 手动。这是我的尝试：

JSON is a very simple format, so it should not be hard to create the convertJsonFormat by hand. Here's my attempt:

static JsonNode convertJsonFormat(JSONObject json) {
    ObjectNode ret = JsonNodeFactory.instance.objectNode();

    @SuppressWarnings("unchecked")
    Iterator<String> iterator = json.keys();
    for (; iterator.hasNext();) {
        String key = iterator.next();
        Object value;
        try {
            value = json.get(key);
        } catch (JSONException e) {
            throw new RuntimeException(e);
        }
        if (json.isNull(key))
            ret.putNull(key);
        else if (value instanceof String)
            ret.put(key, (String) value);
        else if (value instanceof Integer)
            ret.put(key, (Integer) value);
        else if (value instanceof Long)
            ret.put(key, (Long) value);
        else if (value instanceof Double)
            ret.put(key, (Double) value);
        else if (value instanceof Boolean)
            ret.put(key, (Boolean) value);
        else if (value instanceof JSONObject)
            ret.put(key, convertJsonFormat((JSONObject) value));
        else if (value instanceof JSONArray)
            ret.put(key, convertJsonFormat((JSONArray) value));
        else
            throw new RuntimeException("not prepared for converting instance of class " + value.getClass());
    }
    return ret;
}

static JsonNode convertJsonFormat(JSONArray json) {
    ArrayNode ret = JsonNodeFactory.instance.arrayNode();
    for (int i = 0; i < json.length(); i++) {
        Object value;
        try {
            value = json.get(i);
        } catch (JSONException e) {
            throw new RuntimeException(e);
        }
        if (json.isNull(i))
            ret.addNull();
        else if (value instanceof String)
            ret.add((String) value);
        else if (value instanceof Integer)
            ret.add((Integer) value);
        else if (value instanceof Long)
            ret.add((Long) value);
        else if (value instanceof Double)
            ret.add((Double) value);
        else if (value instanceof Boolean)
            ret.add((Boolean) value);
        else if (value instanceof JSONObject)
            ret.add(convertJsonFormat((JSONObject) value));
        else if (value instanceof JSONArray)
            ret.add(convertJsonFormat((JSONArray) value));
        else
            throw new RuntimeException("not prepared for converting instance of class " + value.getClass());
    }
    return ret;
}

注意，杰克逊的 JsonNode 可以代表一些额外的类型（例如 BigInteger ， Decimal 等），因为它们不是必需的上面的代码涵盖 JSONObject 可以表示的所有内容。

Note that, while the Jackson's JsonNode can represent some extra types (such as BigInteger, Decimal, etc) they are not necessary since the code above covers everything that JSONObject can represent.

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