Question

也许有人可以告诉我我在做什么错，我正在编写一个Android应用程序，该应用程序将按时间间隔显示3个视图，因此基本思想是；

    ViewOne.setVisibility(View.INVISIBLE); //This is redundant but I put here for clarity
    ViewTwo.setVisibility(View.INVISIBLE);
    ViewThree.setVisibility(View.INVISIBLE);

    ViewOne.setVisibility(View.VISIBLE);
    //This supposed to make a 1 second stop.
    try {
      Thread.sleep(1000);
    } catch (InterruptedException e) {
      e.printStackTrace();
    }

    ViewTwo.setVisibility(View.VISIBLE);
    //This supposed to make another 1 second stop.
    try {
      Thread.sleep(1000);
    } catch (InterruptedException e) {
      e.printStackTrace();
    }
    ViewThree.setVisibility(View.VISIBLE);

但不是停止单个1000毫秒，而是活动等待2000开始，然后同时显示所有视图。我是android和Java开发人员中的新手，如果我做的是愚蠢的事情，对不起。预先感谢大家。

Answer 1

根本原因：：您的代码阻止了UI线程，因此直到线程空闲（或未阻止）后，它才会更新UI。

解决方案：：我为您提供了此解决方案，可以执行您想要的操作

final Handler handler = new Handler();

ViewOne.setVisibility(View.INVISIBLE); //This is redundant but I put here for clarity
ViewTwo.setVisibility(View.INVISIBLE);
ViewThree.setVisibility(View.INVISIBLE);

ViewOne.setVisibility(View.VISIBLE);
handler.postDelayed(new Runnable() {
    @Override
    public void run() {
        ViewTwo.setVisibility(View.VISIBLE);
        handler.postDelayed(new Runnable() {
            @Override
            public void run() {
                ViewThree.setVisibility(View.VISIBLE);
            }
        }, 1000);
    }
}, 1000);

确保在final和ViewTwo变量之前添加ViewThree关键字。

更新：我不知道您背后的逻辑代码或预期的逻辑代码，但是如果您想重复N次。

int secondsForEachStep = 3;
for (int i = 0; i < 3; i++) {
    handler.postDelayed(new Runnable() {
        @Override
        public void run() {
            wordText.setVisibility(View.INVISIBLE);
            myImageView.setVisibility(View.INVISIBLE);
            handler.postDelayed(new Runnable() {
                @Override
                public void run() {
                    wordText.setVisibility(View.VISIBLE);
                    handler.postDelayed(new Runnable() {
                        @Override
                        public void run() {
                            myImageView.setVisibility(View.VISIBLE);
                        }
                    }, 1000);
                }
            }, 1000);

        }
    }, (i * secondsForEachStep + 1) * 1000);
}

解释以上代码：

i = 0: After 1 seconds hide 2 views, after 2 seconds show view one, after 3 seconds show view 2.
i = 1: After 4 seconds hide 2 views, after 5 seconds show view one, after 6 seconds show view 2.
i = 2: After 7 seconds hide 2 views, after 8 seconds show view one, after 9 seconds show view 2.

对于每个步骤，我们需要3秒才能完成，由此我们可以发现规则是

i * secondsForEachStep + 1

同一Android Activity中的两个Thread.sleep（）

1 个答案: