Question

我为文件浏览器创建了一个新线程。该线程读取目录的内容。我想要做的是更新UI线程以绘制文件和文件夹的图形表示。我知道我无法在新线程中更新UI，所以我想要做的是：

虽然文件扫描线程遍历目录文件，文件夹将文件路径字符串传递回UI线程。然后，UI线程中的处理程序将绘制传回的文件的图形表示。

public class New_Project extends Activity implements Runnable {

    private Handler handler = new Handler() {
        @Override
        public void handleMessage(Message msg) {
            Log.d("New Thread","Proccess Complete.");
            Intent intent = new Intent();
            setResult(RESULT_OK, intent);
            finish();
        }
    };


    public void fileScanner(){
        //if (!XMLEFunctions.canReadExternal(this)) return;
        pd = ProgressDialog.show(this, "Reading Directory.",
                "Please Wait...", true, false);

        Log.d("New Thread","Called");
        Thread thread = new Thread(this);
        thread.start();
    }


    public void run() {
        Log.d("New Thread","Reading Files");
        getFiles();
        handler.sendEmptyMessage(0);
    }


    public void getFiles() {
        for (int i=0;i<=allFiles.length-1;i++){
            //I WANT TO PASS THE FILE PATH BACK TU A HANDLER IN THE UI
            //SO IT CAN BE DRAWN.
            **passFilePathBackToBeDrawn(allFiles[i].toString());**
        } 
    }

}

Answer 1

似乎传递简单的消息是基于int的...我需要做的是传递一个Bundle

using Message.setData(Bundle) and Message.getData(Bundle)

快乐= 0）

//Function From Within The Thread

public void newProjectCreation() {

Message msg =  new Message();
Bundle bundle = new Bundle();

bundle.putString("Test", "test value");
msg.setData(bundle);

handler2.sendMessage(msg);
}

//Handler in The UI Thread Retreieves The Data
//And Can Update the GUI as Required

private Handler handler2 = new Handler() {
    @Override
    public void handleMessage(Message msg) {

    Bundle bundle = msg.getData();
    Toast.makeText(New_Project.this,bundle.getString("Test"),Toast.LENGTH_SHORT).show();

}

};

Answer 2

查看AsyncTask以获取此类内容。它比滚动自己的处理程序和来回传递消息要优雅得多。

http://developer.android.com/reference/android/os/AsyncTask.html

将消息从线程传递到更新UI

2 个答案: