Java 安卓-CharStyle_xtview不会立即更改
我从服务器接受数组,然后填充数据textviewJava 安卓-CharStyle_xtview不会立即更改,java,android,textview,Java,Android,Textview,我从服务器接受数组,然后填充数据textview private class ReadMessages extends AsyncTask<Void, Void, Integer> { HttpURLConnection conn; String ansver, bfr_st; JSONArray JsonArray; protected Integer doInBackground(Void... params) { tr
private class ReadMessages extends AsyncTask<Void, Void, Integer>
{
HttpURLConnection conn;
String ansver, bfr_st;
JSONArray JsonArray;
protected Integer doInBackground(Void... params) {
try {
String post_url = server_name +"/chat.php?action=select";
conn = (HttpURLConnection) new URL(post_url).openConnection();
conn.setReadTimeout(10000);
conn.setConnectTimeout(15000);
conn.setRequestMethod("GET");
conn.setRequestProperty("User-Agent", "Mozilla/5.0");
conn.setDoInput(true);
conn.connect();
InputStream is = conn.getInputStream(); //канал
BufferedReader br = new BufferedReader(new InputStreamReader(is, "UTF-8")); //буфер
StringBuilder sb = new StringBuilder(); //сборщик строки
while ((bfr_st = br.readLine()) != null) {
sb.append(bfr_st); //получили массив в виде string
}
String json_bum = sb.toString();
JsonArray = new JSONArray(json_bum); //преобразовали string обратно в массив
for (int i=0; i<JsonArray.length(); i++) {
jsonObject = JsonArray.getJSONObject(i); //вынули все обьекты
}
tv_number.setText("Номер - " + jsonObject.getInt("_id") +
"\n"
+ "Автор - " + jsonObject.getString("author") +
"\n"
+ "Адресат - " + jsonObject.getString("client") +
"\n"
+ "Время - " + jsonObject.getLong("data") +
"\n"
+ "Текст - " + jsonObject.getString("text") +
"\n");
is.close();
br.close();
} catch (Exception e) {
} finally {
conn.disconnect();
}
return null;
}
}
当我折叠应用程序,然后返回到它的文本变化。但当您按下一个没有发生的按钮时,您只能从应用程序主线程更新GUI元素/小部件。AsyncTask正在创建单独的线程,您无法从方法doInBackground执行此操作,请尝试使用publishProgress和onProgressUpdate,或者在onPostExecute中执行所有这些操作。为什么您决定必须立即执行?这取决于您的互联网速度、ping等。好的做法是使用progressBar向用户显示正在发生的事情,而不仅仅是空屏幕,几秒钟后出现textView。不允许您在单独的线程上更新GUI。您应该在AsyncTask中的OnPostExecute中更新它。此外,您必须@Override doInBackground
doInbackground中的行:
tv_number.setText("Номер - " + jsonObject.getInt("_id") +
"\n"
+ "Автор - " + jsonObject.getString("author") +
"\n"
+ "Адресат - " + jsonObject.getString("client") +
"\n"
+ "Время - " + jsonObject.getLong("data") +
"\n"
+ "Текст - " + jsonObject.getString("text") +
"\n");
实际上抛出一个异常,但是空的catch语句防止应用程序崩溃
不能从后台线程更新视图,更新UI必须在UI线程上。对于AsyncTask,您可以在onPostExecute或onProgressUpdate方法中执行此操作
更改doInBackground实现,使其返回字符串:
protected String doInBackground(Void... params){
}
然后在onPostExecute中,更新textview
protected onPostExecute(String result){
tv_number.setText(result);
}
您正在调用服务器,所以您应该等到它有响应,这取决于许多不同的因素,认为这将立即发生
protected onPostExecute(String result){
tv_number.setText(result);
}