Android发布，然后响应
Android发布，然后响应
提问于 2016-04-19 18:37:58
我可以通过JSON在listview中检索数据，但是我希望显示与登录的用户相关的数据，因此我必须将用户名发布到PHP脚本。我不知道如何将用户名发布到PHP脚本，然后从web服务器获得响应。
private class GetFixture extends AsyncTask<Void, Void, Void> {
@Override
protected void onPreExecute() {
super.onPreExecute();
}
@Override
protected Void doInBackground(Void... arg) {
ServiceHandler serviceClient = new ServiceHandler();
Log.d("url: ", "> " + URL_ITEMS);
String json = serviceClient.makeServiceCall(URL_ITEMS,ServiceHandler.GET);
// print the json response in the log
Log.d("Get match fixture resps","> " + json);
if (json != null) {
try {
Log.d("try", "in the try");
JSONObject jsonObj = new JSONObject(json);
Log.d("jsonObject", "new json Object");
// Getting JSON Array node
matchFixture = jsonObj.getJSONArray(TAG_FIXTURE);
Log.d("json aray", "user point array");
int len = matchFixture.length();
Log.d("len", "get array length");
for (int i = 0; i < matchFixture.length(); i++) {
JSONObject c = matchFixture.getJSONObject(i);
String matchId = c.getString(TAG_MATCHID);
Log.d("matchId", matchId);
String teamA = c.getString(TAG_TEAMA);
Log.d("teamA", teamA);
String teamB = c.getString(TAG_TEAMB);
Log.d("teamB", teamB);
// hashmap for single match
HashMap<String, String> matchFixture = new HashMap<String, String>();
// adding each child node to HashMap key => value
matchFixture.put(TAG_MATCHID, matchId);
matchFixture.put(TAG_TEAMA, teamA);
matchFixture.put(TAG_TEAMB, teamB);
matchFixtureList.add(matchFixture);
}
}
catch (JSONException e) {
Log.d("catch", "in the catch");
e.printStackTrace();
}
} else {
Log.e("JSON Data", "Didn't receive any data from server!");
}
return null;
}
@Override
protected void onPostExecute(Void result) {
super.onPostExecute(result);
ListAdapter adapter = new SimpleAdapter(
Doctor_Names.this, matchFixtureList,
R.layout.list_item, new String[] {
TAG_MATCHID, TAG_TEAMA,TAG_TEAMB
}
, new int[] {
R.id.teamA,R.id.name,
R.id.teamB
}
);
setListAdapter(adapter);
}主席先生，这段代码完美地显示了来自这个服务器的所有数据，但是我想要一些特定的数据，比如我想显示与登录的人相关的数据，所以我必须将用户名传递给PHP脚本，这样我就不会想到POST任何可以过滤数据的东西了。
回答 1
Stack Overflow用户
发布于 2016-04-19 19:02:15
这里是一个示例
JSONObject obj = new JSONObject();
obj.put("username", "YourUser");
HttpUrlConnectionJson.sendHTTPData("http://" + serverAddress + "/api/SomeUserMethod", obj);HttpUrlConnectionJson类
public class HttpUrlConnectionJson {
private static final String TAG = "HttpUrlConnectionJson";
public static String sendHTTPData(String urlpath, JSONObject json) {
HttpURLConnection connection = null;
try {
URL url=new URL(urlpath);
connection = (HttpURLConnection) url.openConnection();
connection.setDoOutput(true);
connection.setDoInput(true);
connection.setRequestMethod("POST");
connection.setRequestProperty("Content-Type", "application/json");
connection.setRequestProperty("Accept", "application/json");
OutputStreamWriter streamWriter = new OutputStreamWriter(connection.getOutputStream());
streamWriter.write(json.toString());
streamWriter.flush();
StringBuilder stringBuilder = new StringBuilder();
if (connection.getResponseCode() == HttpURLConnection.HTTP_OK){
InputStreamReader streamReader = new InputStreamReader(connection.getInputStream());
BufferedReader bufferedReader = new BufferedReader(streamReader);
String response = null;
while ((response = bufferedReader.readLine()) != null) {
stringBuilder.append(response + "\n");
}
bufferedReader.close();
Log.d(TAG, stringBuilder.toString());
return stringBuilder.toString();
} else {
Log.e(TAG, connection.getResponseMessage());
return null;
}
} catch (Exception exception){
Log.e(TAG, exception.toString());
return null;
} finally {
if (connection != null){
connection.disconnect();
}
}
}
}我希望这能帮上忙
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/36726784
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