我知道这是一个重复的问题,但请坚持下去。我已经阅读了一些类似的问题和答案,但似乎没有一个对我有用。
怎么做:
我必须进行搜索,它将向Web服务发送请求并收到响应。
由于我无法在UI线程上使用网络,因此我使用了AsyncTask
。
我尝试了什么:
我尝试使用task.execute()
立即返回,甚至没有显示progressdialog框,我收到响应为null(在onPostExecute
中设置)
如果我使用task.execute.get()
,它会冻结屏幕并再次显示没有对话框(但我收到的回复正确)。
下面是我的task.execute代码。请指正。
public class LookIn extends AppCompatActivity implements View.OnClickListener{
private Button btn=null;
private TextView txtPinCode=null;
private Service service=null;
private final static int timeout=20;
private String jsonResponse;
//private ProgressBar helperSearchProgressBar;
private String pincode="";
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_look_in);
btn=(Button)findViewById(R.id.button);
btn.setOnClickListener(this);
txtPinCode=(TextView) findViewById(R.id.txtPinCode);
this.service=(Service) ParamFactory.getParam(ConstantLabels.SELECTED_SERVICE_ID);
// this.helperSearchProgressBar=(ProgressBar)findViewById(R.id.helperSearchProgressBar);
}
@Override
public void onClick(View v) {
String pincode= txtPinCode.getText().toString();
if(pincode==null || pincode.isEmpty() || pincode.length()!=6)
{
this.txtPinCode.setError("Please enter a 6 degit pin code from 700000 to 700200");
return;
}
ParamFactory.setParam(ConstantLabels.PINCODE_ID,pincode);
this.pincode=pincode;
loadHelper();
Intent intent= new Intent(LookIn.this,SearchResult.class);
startActivity(intent);
}
public void setJsonResponse(String jsonResponse)
{
this.jsonResponse=jsonResponse;
}
private void loadHelper()
{
Log.v("Callme", "Running thread:" + Thread.currentThread().getId());
ArrayAdapter<User> adapter=null;
String params=this.pincode+","+this.service.getId();
List<User> result=null;
try {
new CallmeGetHelperAsyncTask().execute(params); //my task.execute()
result= RestUtil.getUserList(jsonResponse);
adapter = new ArrayAdapter(this, android.R.layout.simple_list_item_1, result);
ParamFactory.setParam("getHelperForService", adapter);
}
catch(JSONException x)
{
Log.e("Callme", Log.getStackTraceString(x));
}
}
class CallmeGetHelperAsyncTask extends AsyncTask<String,Void,String > {
// private Context context=null;
private ProgressDialog dialog=null;
private String jsonResponse;
private LookIn activity;
public CallmeGetHelperAsyncTask(){}
public CallmeGetHelperAsyncTask(LookIn activity)
{
this.activity=activity;
}
@Override
protected void onPreExecute() {
this.dialog= new ProgressDialog(LookIn.this);
this.dialog.setMessage("Loading...");
this.dialog.show();
Log.v("Callme","Dialog Shown");
}
@Override
protected void onPostExecute(String s) {
if(s!=null)
{
this.activity.setJsonResponse(s);
}
else
{
Log.v("Callme","kill me");
}
if(this.dialog.isShowing())
{
Log.v("Callme","Closing Dialog");
this.dialog.dismiss();
}
}
@Override
protected String doInBackground(String... params) {
Log.v("Callme","From Background:"+Thread.currentThread().getId());
String pincode=params.clone()[0].split(",")[0];
String serviceId=params.clone()[0].split(",")[1];
String url=String.format(URL.GET_HELPER,serviceId,pincode);
jsonResponse= null;
try {
jsonResponse = RestUtil.makeRestRequest(url);
} catch (IOException e) {
e.printStackTrace();
}
return jsonResponse;
}
}
}
注意:我没有尝试使用while循环来等待asynctask,因为我认为这也会冻结我的屏幕。如果我错了,请纠正我
答案 0 :(得分:3)
我没有尝试使用while循环来等待asynctask
无需使用循环来等待AsyncTask
结果。
因为onPostExecute
方法在doInBackground
之后执行,所以不是在调用jsonResponse
方法后使用execute
,而是在setJsonResponse
方法内执行,因为此方法从onPostExecute
调用,始终在主UI线程上运行:
public void setJsonResponse(String jsonResponse)
{
this.jsonResponse=jsonResponse;
//Create adapter object here
result= RestUtil.getUserList(jsonResponse);
adapter = new ArrayAdapter(...);
ParamFactory.setParam("getHelperForService", adapter);
}