拜托,请帮帮我..
我正在开发一个项目,我从 JSON格式的Web服务获取数据.我试图解析它,但我无法做到.我有这个json数据 –
我正在开发一个项目,我从 JSON格式的Web服务获取数据.我试图解析它,但我无法做到.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"
}
]
}
]
}
}
从这里我成功地获得了“食物”阵列,但当我试图获得“内容”数组和food_name数据时,我陷入困境.
我正在使用此代码,但我没有得到任何解决方案,请检查此剪辑代码.
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
}
}
检查此网址以获取网络服务响应 –
WEB-SERVICE URL
解决方法
您需要使用以下代码替换相应的代码部分：
for (int j = 0; j < FoodNameArray.length(); j++) {
String Foodname = FoodNameArray.getJSONObject(j).getString("food_name");
System.out.println("@@@@@@@@@@@@@" + Foodname);
}