我正在尝试解析一个
JSON对象,其中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,通过这样做,我能够非常简单地解析：
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’名称’时,它将无法正常工作.
我试过了：
JSONObject businessObject = jObject.getJSONObject("business");
String nameValue = businesObject.getString("name");
System.out.println(nameValue);
当我尝试这个时,我得到“JSONObject [business] not found.”
当我尝试：
String nameValue = offerObject.getString("name");
System.out.println(nameValue);`
正如预期的那样,我得到“未找到JSONObject [name]”.
我在这做错了什么？我遗漏了一些基本的东西……
解决方法
好吧,我是个白痴.这很有效.
JSONObject businessObject = offerObject.getJSONObject("business");
String nameValue = businessObject.getString("name");
System.out.println(nameValue);
如果我只想在张贴前两秒钟……杰斯！
