我不明白为什么我得到这个错误。我使用的AsyncTask运行在后台的一些过程。

I don't understand why I'm getting this error. I'm using AsyncTask to run some processes in the background.

我有：

protected void onPreExecute() 
{
    connectionProgressDialog = new ProgressDialog(SetPreference.this);
    connectionProgressDialog.setCancelable(true);
    connectionProgressDialog.setProgressStyle(ProgressDialog.STYLE_SPINNER);
    connectionProgressDialog.setMessage("Connecting to site...");
    connectionProgressDialog.show();

    downloadSpinnerProgressDialog = new ProgressDialog(SetPreference.this);
    downloadSpinnerProgressDialog.setProgressStyle(ProgressDialog.STYLE_SPINNER);
    downloadSpinnerProgressDialog.setMessage("Downloading wallpaper...");
}

当我进入 doInBackground（）根据条件我：

[...]    
connectionProgressDialog.dismiss();
downloadSpinnerProgressDialog.show();
[...]

每当我尝试 downloadSpinnerProgressDialog.show（）我收到错误。

任何想法家伙？

推荐答案

方法显示（）必须从的用户界面（UI）线程，而 doInBackground（）运行在不同的线程这是最主要的原因的AsyncTask 设计。

The method show() must be called from the User-Interface (UI) thread, while doInBackground() runs on different thread which is the main reason why AsyncTask was designed.

您必须调用显示（）无论是在 onProgressUpdate（）或 onPostExecute （）。

You have to call show() either in onProgressUpdate() or in onPostExecute().

例如：

class ExampleTask extends AsyncTask<String, String, String> {

    // Your onPreExecute method.

    @Override
    protected String doInBackground(String... params) {
        // Your code.
        if (condition_is_true) {
            this.publishProgress("Show the dialog");
        }
        return "Result";
    }

    @Override
    protected void onProgressUpdate(String... values) {
        super.onProgressUpdate(values);
        connectionProgressDialog.dismiss();
        downloadSpinnerProgressDialog.show();
    }
}