I have a function, AppHelper.isOnline(Context context)
, I call in various parts of my application to check that a session didn't timeout before making an HTTP request.
public void onClick(View v) {
Intent intent = null;
switch (v.getId()) {
case R.id.buttonPagamenti:
if (AppHelper.isOnline(this))
{
//here AppHelper.isOnline should have finished it's async task
intent = new Intent(this, OrdineCreaActivity.class);
this.startActivityForResult(intent, R.id.buttonPagamenti);
}
break;
...
Inside AppHelper.isOnline()
, I am executing an AsyncTask
that logs in, thus making a network request, which can't be run on UI because otherwise I get an exception. I need to wait for it to finish BEFORE resuming with the code inside the if. How can I do this ?
Problem is the activity starts firsts, then the AsyncTask
executes, so when the activity expects a valid logged in session, it breaks.
You have two options:
Either use the AsyncTask
's method get(long timeout, TimeUnit unit)
like that:
task.get(1000, TimeUnit.MILLISECONDS);
This will make your main thread wait for the result of the AsyncTask
at most 1000 milliseconds (as per @user1028741 comment: actually there is also infinetly waiting method - AsyncTask#get()
which might also do the work for you in some cases).
Alternatively you can show a progress dialog in the async task until it finishes. See this thread (No need for me to copy past the code). Basically a progress dialog is shown while the async task runs and is hidden when it finishes.
You have even third option:" if Thread
is sufficient for your needs you can just use its join
method. However, if the task is taking a long while you will still need to show a progress dialog, otherwise you will get an exception because of the main thread being inactive for too long.