Question

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.

Answer 1

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.