Question

I m kinda new to Android, now I m developing a very simple game, the logic is described as following: The user sees a 10-digit value after pressing a "Ready" Button. After 5 seconds, the value changes to
* * * * * * * and the user has to enter it in a text field (type "number") below. Under that text field there are 2 Buttons: "Check" and "Give a hint". The check Button compares the user value to original value and changes a TextView according to the result("Correct"/"Incorrect"). The hint Button should show 3 random digits in the original value.

我对这项小应用有一些问题:

I m not sure which type I have to use to show the original value to the user: a TextView, a text field or something else?
How do I force the app to show 3 random digits from the original value in the case when user presses the hint Button?

感谢任何帮助。

Answer 1

页: 1 我不敢确定哪类我必须用来向用户展示原始价值:文本意见、案文领域或其他内容?

本案文将照此办理。

页: 1 当用户对 h子施加压力时,我如何迫使该数字显示3位随机数字?

希望你利用int储存10位数值(原始价值)

1. 形成规模10的阵容

int[] digits = new int[10];

用于将10位数与10位数分开。

int number = 1234567891
for(int i = 0; i < 10; i++){
  digits[i] = number % 2;
  number = number / 10;
}

这将赋予你10个内在价值。

object。随机值从0-9

Random r = new Random();
int pos1 = r.nextInt(9);

nextInt (int n) Return a pseudo-random coordinatedly distribution int in the half-open range [0, n].

These values are positions from which you are going to retrieve digit from array So this will give you random position between 0 and 9

如果你为第二位数生成随机位置,你应核实这种位置应等于第一。

you can show all 3 digits(fetched from 3 randomly generated positions) in TextView as a hint Hope this will solve your problem

Edited Part Set visibility of Textview to View.GONE something like this private TextView tv;

tv = (TextView)findViewById(R.id.textView01);
tv.setVisibility(View.GONE);

Button btn = (Button)findViewById(R.id.button01);
btn.setOnClickListener(new OnClickListener(){
   public void onClick(View view){
      tv.setVisibility(View.VISIBLE);
   }
});

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