Question

我正在开发一个将英语单词翻译成波斯语的应用程序当我把小数据库放在每个东西都顺利的时候。但当我使用长数据库时UI冻结。然后我使用Asynchronous在后台加载数据库，但我不能将执行方法的结果发送到Activity中的List View？

public class MainActivity extends ListActivity {

private database db;
private EditText ed_txt;
private TextView tv;
private RadioButton rb_en;
private RadioButton rb_fa;

private String[] searched_word;
private String[] en;
private String[] fa;



@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    db=new database(this, "dictionary", null, 1);
    db.useable();

    ed_txt=(EditText)findViewById(R.id.ed_txt);
    tv=(TextView)findViewById(R.id.tv);
    rb_en=(RadioButton)findViewById(R.id.rb_en);
    rb_fa=(RadioButton)findViewById(R.id.rb_fa);

    //refresher(ed_txt.getText().toString(), "en");

    ed_txt.addTextChangedListener(new TextWatcher() {


        @Override
        public void onTextChanged(CharSequence arg0, int arg1, int arg2, int arg3) {
            if(rb_en.isChecked()){
            refresher(ed_txt.getText().toString(), "en");

        }
        else if (rb_fa.isChecked()) {
            refresher(ed_txt.getText().toString(), "per");

        }
        }

        @Override
        public void beforeTextChanged(CharSequence arg0, int arg1, int arg2,
                int arg3) {
            // TODO Auto-generated method stub

        }

        @Override
        public void afterTextChanged(Editable arg0) {
            // TODO Auto-generated method stub
             new AsynctaskClass(handler).execute();
        }
    });



}

public class AsynctaskClass extends AsyncTask<Void, Void, Void> {

    private Handler handler;
    public AsynctaskClass(Handler handler ){
     this.handler = handler;
   }
    @Override
    protected Void doInBackground(Void... params) {
       Message message = handler.obtainMessage();
       //response you wish to send to the listview as String object
       message.obj =new String();
       message.what = 1 ;//You can use any value which will help you to distinguish the Handler response received at the activity.
       handler.sendMessage(message);
    return null;

    }

}

Handler handler = new Handler() {
    @Override
    public void handleMessage(Message msg) {
        super.handleMessage(msg);
    }
};


class AA extends ArrayAdapter<String>{

    public AA(){
        super(MainActivity.this,R.layout.row_search,en);
    }

    @Override
    public View getView(final int position, View convertView, ViewGroup parent) {
        LayoutInflater in=getLayoutInflater();
        View row=in.inflate(R.layout.row_search, parent, false);


        TextView tv_searched_word=(TextView)row.findViewById(R.id.tv_searched_word);

        tv_searched_word.setText(en[position]);

        return (row);
    }

}


private void refresher(String text, String field){
    db.open();
    int s = db.shmaresh_jostojoo(text, field);
    if (ed_txt.getText().toString().equals("")) {
        s = 0;
        tv.setText(" لطفا کلمه مورد نطرتان را وارد کنید");
    }else {
        tv.setText(" تعداد "+s+" یافت شد ");

    }
    //searched_word[s];
    en=new String[s];
    fa=new String[s];

    for (int i = 0; i < s; i++) {
        //searched_word[i]=db.jostojoo(i, col, word, field);
        if(field.equals("en")){
            en[i]=db.jostojoo(i, 1, text, field);
        }else{
            en[i]=db.jostojoo(i, 2, text, field);
        }
        //en[i]=db.jostojoo(i, 1, text, field);
        //fa[i]=db.jostojoo(i, 2, text, field);
    }



    setListAdapter(new AA());
    db.close();
}

https://github.com/mohammadi66/Dictionary-En-2-Fa.git

Answer 1

在你的代码中，我看不到任何asynctask，但使用Handler可以解决你的问题。

在您的Activity中定义处理程序如下：

Handler handler = new Handler() {
                    @Override
                    public void handleMessage(Message msg) {
                        super.handleMessage(msg);
                    }
 }

然后使用此处理程序对象并将其传递给asynctask，如下所示：

  new AsynctaskClass(handler, other params as you need).execute();

现在你的asynctask看起来如下：

public class AsynctaskClass extends AsyncTask<Void, Void, Void> {

    private Handler handler;
    public AsynctaskClass(Handler handler, other params ){
     this.handler = handler;
   }
   ...
   ...
}

现在在Asynctask的 doInBackground（）中，执行您喜欢的任何操作，然后当您获得要发送到listview的结果时，添加以下代码：

@Override
protected Void doInBackground(Void... params) {
   Message message = handler.obtainMessage();
   message.obj = response you wish to send to the listview as String object;
   message.what = 1 //You can use any value which will help you to distinguish the Handler response received at the activity.
   handler.sendMessage(message);

}

就是这样..现在当sendMessage()将被调用asynctask ...在你的Acitivity中 handleMessage（）将获得触发器..，因为你可以获得从asynctask as msg.obj.toString() ..现在使用此数据执行UI修改。

希望这会对你有所帮助。

从asynctask获取结果列表视图

1 个答案: