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。
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!