在Android中解析嵌套的JSON对象

Question

我正在尝试解析JSON对象，其中的一部分看起来像这样：

{
"offer":{
    "category":"Salon",
    "description":"Use this offer now to enjoy this great Salon at a 20% discount. ",
    "discount":"20",
    "expiration":"2011-04-08T02:30:00Z",
    "published":"2011-04-07T12:00:33Z",
    "rescinded_at":null,
    "title":"20% off at Jun Hair Salon",
    "valid_from":"2011-04-07T12:00:31Z",
    "valid_to":"2011-04-08T02:00:00Z",
    "id":"JUN_HAIR_1302177631",
    "business":{
        "name":"Jun Hair Salon",
        "phone":"2126192989",
        "address":{
            "address_1":"12 Mott St",
            "address_2":null,
            "city":"New York",
            "cross_streets":"Chatham Sq & Worth St",
            "state":"NY",
            "zip":"10013"
        }
    },

等等....

到目前为止，通过执行以下操作，我可以非常简单地解析：

JSONObject jObject = new JSONObject(content);
JSONObject offerObject = jObject.getJSONObject("offer");
String attributeId = offerObject.getString("category");
System.out.println(attributeId);

String attributeValue = offerObject.getString("description");
System.out.println(attributeValue);

String titleValue = offerObject.getString("title");
System.out.println(titleValue);`

但是，当我尝试将其用作“名称：”时，它将不起作用。

我试过了：

JSONObject businessObject = jObject.getJSONObject("business");
String nameValue = businesObject.getString("name");
System.out.println(nameValue);

当我尝试这样做时，我得到“找不到JSONObject [业务]”。

当我尝试：

String nameValue = offerObject.getString("name");
System.out.println(nameValue);`

我得到了预期的“找不到JSONObject [名称]”。

我在这里做错了什么？我缺少一些基本的东西。

Answer 1

好吧，我是个白痴。这可行。

JSONObject businessObject = offerObject.getJSONObject("business");
String nameValue = businessObject.getString("name");
System.out.println(nameValue);

如果我只想想两秒钟，然后再发布…哎呀！