Question

在这里花了几个小时计算如何改变生成数字的颜色

for (int rolls = 0; rolls < 4 +temp; rolls++) {
            value = (int) (Math.random() * 4 + 1);
            //Test if numbers generated are correct
            builder.append(value + " ");
            textTest.setText(builder.toString());               
            NumbersArray[rolls] = value;
            System.out.println(NumbersArray[rolls]);

我设法做到了，现在，我有一个问题，他们所有人同时改变颜色，我怎么能这样做所以生成的第一个数字首先亮起，等待几秒然后下一个生成的数字改变颜色？也许我错了，因为即使没有生成颜色，所有颜色都会改变

            if (rolls == 1) {
                red.setBackgroundColor(Color.parseColor("#FF0000"));
                //Wait t amount of time here
                red.postDelayed(new Runnable() {
                    public void run() {
                        red.setBackgroundColor(Color.parseColor("#FF7D0000")); /* Changes Color  */
                    }
                },3*100 /*  millisecond  to wait */);
            }
            else if (rolls == 2) {
                yellow.setBackgroundColor(Color.parseColor("#FFF000"));
                yellow.postDelayed(new Runnable() {
                    public void run() {
                        yellow.setBackgroundColor(Color.parseColor("#FF7F7500"));
                    }
                }, 3 * 100);
            }
            else if (rolls == 3) {
                green.setBackgroundColor(Color.parseColor("#0FFF00"));
                green.postDelayed(new Runnable() {
                    public void run() {
                        green.setBackgroundColor(Color.parseColor("#FF087D00"));
                    }
                }, 3 * 100 );
            }
            else  {
                blue.setBackgroundColor(Color.parseColor("#0000FF"));
                blue.postDelayed(new Runnable() {
                    public void run() {
                        blue.setBackgroundColor(Color.parseColor("#FF00007D"));
                    }
                }, 3 * 100 );
            }

整个循环：

 for (int rolls = 0; rolls < 2 +temp; rolls++) {
            value = (int) (Math.random() * 4 + 1);
            builder.append(value + " ");
            textTest.setText(builder.toString());
            NumbersArray[rolls] = value;
            System.out.println(NumbersArray[rolls]);


            if (value == 1) {
                red.setBackgroundColor(Color.parseColor("#FF0000"));
                red.postDelayed(new Runnable() {
                    public void run() {
                        red.setBackgroundColor(Color.parseColor("#FF7D0000"));
                   }
                },3000);
            }
            else if (value == 2) {
                yellow.setBackgroundColor(Color.parseColor("#FFF000"));
                yellow.postDelayed(new Runnable() {
                    public void run() {
                        yellow.setBackgroundColor(Color.parseColor("#FF7F7500"));
                    }
                }, 3000);
            }
            else if (value == 3) {
                green.setBackgroundColor(Color.parseColor("#0FFF00"));
                green.postDelayed(new Runnable() {
                    public void run() {
                        green.setBackgroundColor(Color.parseColor("#FF087D00"));
                    }
                }, 3000 );
            }
            else  {
                blue.setBackgroundColor(Color.parseColor("#0000FF"));
                blue.postDelayed(new Runnable() {
                    public void run() {
                        blue.setBackgroundColor(Color.parseColor("#FF00007D"));
                    }
                }, 3000 );
            }

Answer 1

您可以尝试使用此代码暂停内容更新和检查条件的临时时间。此处理程序将首次运行runnable而不会有任何时间延迟，之后需要1000毫秒才能更新内容并将像循环一样运行。

po[0]

如果两件事情匹配

1 个答案: