请帮助我... 我正在进行一个项目,从Web服务中获取JSON格式的数据。我试图解析它,但是我无法做到。我有这个JSON数据-
{
"response": {
"status": {
"code": "1",
"message": "sucess",
"user_id": "1"
},
"foods": [
{
"char": "A",
"content": [
{
"food_name": "add Malt"
},
{
"food_name": "a la mode"
},
{
"food_name": "Almonds"
}
]
},
{
"char": "Z",
"content": [
{
"food_name": "Zebra Cakes"
},
{
"food_name": "Zucchini, Baby"
},
{
"food_name": "zxc"
}
]
}
]
}
}
我能成功获取“foods”数组,但在尝试获取“content”数组和食品名称数据时遇到了困难。
我正在使用以下代码,但没有找到任何解决方案,请检查此代码片段。
protected String doInBackground(String... args) {
// Building Parameters
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
nameValuePairs.add(new BasicNameValuePair("method","eat_tracking_details"));
nameValuePairs.add(new BasicNameValuePair("uid",userid));
// getting JSON string from URL
JSONObject json = jsonParser.makeHttpRequest(JSONParser.urlname,"GET", nameValuePairs);
//System.out.println("****json*"+json);
if (json != null) {
try {
JSONObject response = json.getJSONObject("response");
JSONObject status = response.getJSONObject("status");
code = status.getString("code");
JSONArray FoodArray = response.getJSONArray("foods");
for (int i = 0; i < FoodArray.length(); i++) {
String character = FoodArray.getJSONObject(i).getString("char");
System.out.println("*****character****************"+character);
JSONArray FoodNameArray = new JSONArray(FoodArray.getJSONObject(i).getString("content"));
System.out.println("====================///////////"+FoodNameArray);
for (int j = 0; j <FoodNameArray.length(); j++) {
String Foodname = FoodArray.getJSONObject(j).getString("food_name");
System.out.println("@@@@@@@@@@@@@"+Foodname);
}
}
} catch (JSONException e) {
// TODO: handle exception
}
}
请查看此网址以获取Web服务响应- WEB-SERVICE URL