我刚刚阅读并测试了AsyncTask，现在我需要知道如何在onPostExecute部分传递多个值。我使用JSON解析器从Web获取值，但是我从JSON获取的值是多个值，我将这些值传递给由每个获取的数据的列标题分隔的数组，这是我应该返回它的部分用于onPostExecute使用。但据我所知，每次运行只能使用一次返回（如果我错了请纠正我）。
到目前为止，这是我的代码：
public class GetInfo extends AsyncTask<String, Void, List<String>>{
private final String TAG = null;
InputStream is = null;
StringBuilder sb=null;
List<String> list = new ArrayList<String>();
@Override
protected List<String> doInBackground(String... url) {
String result = "";
ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
//CONNECT TO DATABASE
try{
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(url[0]);
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity = response.getEntity();
is = entity.getContent();
Log.v(TAG, "connected");
}catch(Exception e){
Log.v(TAG, "run failed");
Log.e("log_tag", "Error in http connection "+e.toString());
}
//BUFFERED READER FOR INPUT STREAM
try{
BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),8);
sb = new StringBuilder();
String line = "0";
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
is.close();
result=sb.toString();
Log.v(TAG, "buffered read");
}catch(Exception e){
Log.v(TAG, "buffered error");
Log.e("log_tag", "Error converting result "+e.toString());
}
//CONVERT JSON TO STRING
try{
Log.v(TAG, result);
JSONArray jArray = new JSONArray(result);
JSONObject json_data=null;
for(int i=0;i<jArray.length();i++){
Log.v(TAG, "loop start");
json_data = jArray.getJSONObject(i);
list.add(json_data.getString("id"));
list2.add(json_data.getString("city"));
Log.v(TAG, "list added");
}
}catch(JSONException e){
Log.v(TAG, "rest failed");
Log.e("log_tag", "Error parsing data "+e.toString());
}
Log.v(TAG, list.toString());
return list; //I also need to return the list2 here
}
@Override
protected void onPostExecute(List<String> result) {
cities = result; //lost in this part hahaha!
showCities();
}
}
好的补充一点，当我只返回一个字符串数组（列表）时，这段代码工作正常，但我现在在第二部分感到困惑。此外，城市在Main类中声明，ShowCities（）仅用于显示。所以我不打扰添加代码。
答案 0 :(得分：5)
你可以做一件事使你的ArrayList成为静态，并在你需要时访问它。
public static List<String> LIST = new ArrayList<String>();
public static List<String> LIST1 = new ArrayList<String>();
public class GetInfo extends AsyncTask<String, Void, List<String>>{
private final String TAG = null;
InputStream is = null;
StringBuilder sb=null;
@Override
protected List<String> doInBackground(String... url) {
String result = "";
ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
//CONNECT TO DATABASE
try{
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(url[0]);
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity = response.getEntity();
is = entity.getContent();
Log.v(TAG, "connected");
}catch(Exception e){
Log.v(TAG, "run failed");
Log.e("log_tag", "Error in http connection "+e.toString());
}
//BUFFERED READER FOR INPUT STREAM
try{
BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),8);
sb = new StringBuilder();
String line = "0";
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
is.close();
result=sb.toString();
Log.v(TAG, "buffered read");
}catch(Exception e){
Log.v(TAG, "buffered error");
Log.e("log_tag", "Error converting result "+e.toString());
}
//CONVERT JSON TO STRING
try{
Log.v(TAG, result);
JSONArray jArray = new JSONArray(result);
JSONObject json_data=null;
for(int i=0;i<jArray.length();i++){
Log.v(TAG, "loop start");
json_data = jArray.getJSONObject(i);
LIST.add(json_data.getString("id"));
LIST1.add(json_data.getString("city"));
Log.v(TAG, "list added");
}
}catch(JSONException e){
Log.v(TAG, "rest failed");
Log.e("log_tag", "Error parsing data "+e.toString());
}
Log.v(TAG, list.toString());
return LIST; //I also need to return the list2 here
}
@Override
protected void onPostExecute(List<String> result) {
cities = result; //lost in this part hahaha!
showCities();
}
}
现在你可以使用你的LIST＆amp; LIST1，如果你想要它。您也可能不需要在DoInBackground中返回arraylist。
希望它会帮助你。
答案 1 :(得分：3)
创建一个包含两个列表的新类。使用该类作为返回类型。