Android ListView:通过json数组循环到列表
这方面的工作已经进行了几个小时了,但我并不确定应该如何对我的测试数组进行排序。我已经尝试了一些不同的选项,这是我的项目中最简单的实现方法。有什么建议或想法可以帮助我克服这个困难吗 来自php的JSON数组Android ListView:通过json数组循环到列表,android,Android,这方面的工作已经进行了几个小时了,但我并不确定应该如何对我的测试数组进行排序。我已经尝试了一些不同的选项,这是我的项目中最简单的实现方法。有什么建议或想法可以帮助我克服这个困难吗 来自php的JSON数组 { "Questions": { "0001": { "Title": "What is Stackoverflow", "Answer": "C", "User": "testUSER",
{
"Questions": {
"0001": {
"Title": "What is Stackoverflow",
"Answer": "C",
"User": "testUSER",
"Date": "0000-00-00",
"Used": "0"
},
"0002": {
"Title": "What is Stackoverflow",
"Answer": "C",
"User": "testUSERb",
"Date": "0000-00-00",
"Used": "1"
},
"0003": {
"Title": "What is Stackoverflow",
"Answer": "C",
"User": "testUSERc",
"Date": "0000-00-00",
"Used": "0"
}
},
"Count-start": 1,
"Count-end": 3
}
已使用:0=未使用1=用户已使用。我将使用它来确定是否向用户显示问题和答案
助手类
public JSONObject query(){
HttpClient client = new DefaultHttpClient();
HttpConnectionParams.setConnectionTimeout(client.getParams(), 10000);
HttpResponse response;
JSONObject json = new JSONObject();
HttpPost post = new HttpPost(QUERY_LINK);
post.setHeader("json", json.toString());
StringEntity se;
try {
se = new StringEntity(json.toString());
se.setContentEncoding(new BasicHeader(HTTP.CONTENT_TYPE,"application/json"));
post.setEntity(se);
response = client.execute(post);
if (response != null) {
InputStream in = response.getEntity().getContent();
a = convertStreamToString(in);
JSONObject check = new JSONObject(a);
return check;
}
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
} catch (JSONException e) {
e.printStackTrace();
}
return json;
}
主要活动
private ArrayList<ListHelper> GetSearchResults(){
JSONObject getJSON = new Helper().query();
JSONArray jArrayObject = new JSONArray();
jArrayObject.put(getJSON);
int end = 0,start = 0;
String title, answer, user, used;
ListHelper sr = new ListHelper();
ArrayList<ListHelper> results = new ArrayList<ListHelper>();
JSONObject offerObject = null;
try {
offerObject = getJSON.getJSONObject("Questions");
start = offerObject.getInt("Count-start");
end = offerObject.getInt("Count-end");
} catch (JSONException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
for(int a = start; a < end; a = a++) {
try {
JSONObject businessObject = offerObject.getJSONObject("Id");
// this is were i'm stuck, how to go about selecting each id
title = businessObject.getString("Title");
answer = businessObject.getString("Answer");
user = businessObject.getString("User");
used = businessObject.getString("Used");
sr = new ListHelper();
sr.setTitle(title);
sr.setUser(user);
sr.setUsed(used);
results.add(sr);
} catch (JSONException e1) {
e1.printStackTrace();
}
}
return results;
}
}
private ArrayList GetSearchResults(){
JSONObject getJSON=new Helper().query();
JSONArray jArrayObject=新的JSONArray();
jArrayObject.put(getJSON);
int end=0,start=0;
字符串标题、答案、用户、已使用;
ListHelper sr=新的ListHelper();
ArrayList结果=新建ArrayList();
JSONObject offerObject=null;
试一试{
offerObject=getJSON.getJSONObject(“问题”);
start=offerObject.getInt(“计数开始”);
end=offerObject.getInt(“计数结束”);
}捕获(JSONException e){
//TODO自动生成的捕捉块
e、 printStackTrace();
}
for(int a=start;a
您可以执行以下操作:在for
循环之外定义一个新的十进制格式
DecimalFormat myFormatter = new DecimalFormat("0000");
然后在for
循环中使用格式化程序形成id
字符串
String id = myFormatter.format(a);
然后使用id
获取JSONObject
// this is were i'm stuck, how to go about selecting each id
JSONObject businessObject = offerObject.getJSONObject(id);
仅当您的ID
最多为4位时,上述操作才有效。您可以使用“Count end”
值而不是使用硬编码的“0000”
使格式动态
如果“Questions”
对象是一个JSONArray,那么所有这些问题都可以避免。来自php的Json数据不是Json数组类型。。它完全是Json对象。