Question

我有一个TextView，我想用我的字符串数组中的每个字符串每5秒更新一次TextView。

这是我试过的代码。它始终只显示String数组中的最后一个String。

TextView display;
EditText caption;
Thread thread;
String blinks;
String[] wc;
private CountDownTimer timer;

@Override
protected void onCreate(Bundle savedInstanceState) {

    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    display = (TextView) findViewById(R.id.display);
    caption = (EditText) findViewById(R.id.caption);

    timer = new CountDownTimer(5000, 20) {

        @Override
        public void onTick(long millisUntilFinished) {

            String[] wc = {"The","Qucik", "Brown","fox","Jumped"};
            for (int j = 0; j < wc.length; j++) {

                blinks = wc[j];
                final String[] titles = {"" + blinks + ""};

                for (int i = 0; i < titles.length; i++) {
                    display.setText(titles[i]);
                }

            }


        }

        @Override
        public void onFinish() {
            try{
                yourMethod();
            }catch(Exception e){

            }
        }
    }.start();
}

Answer 1

final String[] wc = {"The", "Qucik", "Brown", "fox", "Jumped"};
        final android.os.Handler handler = new android.os.Handler();
        handler.post(new Runnable() {

            int i = 0;

            @Override
            public void run() {
                display.setText(wc[i]);
                i++;
                if (i == wc.length) {
                    handler.removeCallbacks(this);
                } else {
                    //5 sec
                    handler.postDelayed(this, 1000 * 5);
                }
            }
        });

Answer 2

尝试在android中使用处理程序，我认为你正在这样：

String[] wc = {"The","Qucik", "Brown","fox","Jumped"};
int j = 0;
Handler handler = new Handler(); 
Runnable runit = new Runnable() { 

    @Override 
    public void run() { 
     try{

            for (; j < wc.length; j++) {

                blinks = wc[j];
                final String[] titles = {"" + blinks + ""};

                for (int i = 0; i < titles.length; i++) {
                    display.setText(titles[i]);
                }
                handler.postDelayed(this,5000);
            }
        }
        catch (Exception e) {
            //  handle exception
        }
    } 
}; 
handler.postDelayed(runit, 5000);

Answer 3

问题：for循环的最后一次迭代（在j上）将文本设置为最后一个字符串。

您需要确保每次打勾只拨打TextView.setText()一次。然后，您可以使用您的方法（CountDownTimer）或Antrrommet's（Handler）。使用您的方法，这将类似于：

int counter = 0; // field member, NOT local variable

@Override
public void onTick(long millisUntilFinished) {
    display.setText("{" + wc[counter++] + "}");
    if (counter == 5) counter = 0;  
}

Answer 4

如果答案有用。请投票：）

        final String[] gritArray = getResources().getStringArray(R.array.gritInformation);
    //creating new handler allows send and process message
        final Handler gritInfoHanlder = new Handler();
    //adding runnable to message queque
            gritInfoHanlder.post(new Runnable() {
    //initializing array position
                int tipPosition = 0;

                @Override
                public void run() {
                    //set number of tip(randon/another way)
//setting array to textview
                    gritInfo.setText(gritArray[tipPosition]);
                    tipPosition++;
//if array exist its length remove callback
                    if (tipPosition == gritArray.length) {
                        gritInfoHanlder.removeCallbacks(this);
//delayed handler 10 sec
                    } else {
                        gritInfoHanlder.postDelayed(this, 1000 * 10);
                    }
                }
            });

每5秒使用String-array中的每个字符串更新TextView

4 个答案: