问题描述
)我正在创建一个记忆游戏.我的问题是，每当我第二次点击时，我什至看不到切换按钮.要清楚 - 第一次单击切换切换按钮，所以我可以看到它持有的数字，第二次单击不同的切换按钮应该切换它，显示数字，然后继续设置分数 +1，如果数字是相同，或者如果它们不同，则再次反转它们.
)I'm in a process of creating a memory game. My problem is that whenever i click for the second time, i can't even see toggled button. To be clear - first click toggles the togglebutton, so i can see the number it holds, the second click on a different togglebutton is suposed to toggle it, show me the number and then proceed to either set a score +1 if numbers are the same, or reverse them back again if they're different.
下面是我用作 onClick 函数的代码，我一直在考虑在第二个if 块"中放置某种睡眠或延迟函数 - (if(klikniecia ==2)).
Below is the code that i use as my onClick function, i've been thinking about putting some kind of sleep or delay function somwhere in the second "if block" - (if(klikniecia ==2)).
任何有关此主题的帮助将不胜感激.
Any help on this topic would be appreciated.
public void onClick(View view) {
for (int i = 0; i < karta.length; i++){
if (view == karta[i]){
karta[i].setEnabled(false);
klikniecia++;
if (klikniecia == 1){
kartaID[0]=i;
kartaWartosc[0]=listaKart.get(i);
}
if (klikniecia == 2){
kartaID[1]=i;
kartaWartosc[1]=listaKart.get(i);
//i think, about setting a delay here, so i can see both of the cards, regardles if the're the same or not before reverting them.
if (czyPara()){
karta[kartaID[0]].setEnabled(false);
karta[kartaID[1]].setEnabled(false);
klikniecia=0;
}
else{
karta[kartaID[0]].setEnabled(true);
karta[kartaID[0]].toggle();
karta[kartaID[1]].setEnabled(true);
karta[kartaID[1]].toggle();
klikniecia=0;
}
}
}
}
}
推荐答案
您可以将延迟消息发布到处理程序(带有关联的可运行对象)并让它更新 UI，而不是在 onclick 中休眠.显然将其融入您的应用设计并使其工作，但基本思想是这样的:
Instead of sleeping in the onclick, you could post a delayed message to a handler (with associated runnable) and have it update the UI. Obviously fit this into the design of your app and make it work, but the basic idea is this:
//Here's a runnable/handler combo
private Runnable mMyRunnable = new Runnable()
{
@Override
public void run()
{
//Change state here
}
};
然后从 onClick 将延迟消息发布到处理程序，指定要执行的可运行对象.
Then from onClick you post a delayed message to a handler, specifying that runnable to be executed.
Handler myHandler = new Handler();
myHandler.postDelayed(mMyRunnable, 1000);//Message will be delivered in 1 second.
根据您的游戏的复杂程度，这可能不是您想要的，但它应该让您有一个开始.
Depending on how complicated your game is, this might not be exactly what you want, but it should give you a start.
这篇关于如何在 Android onClick 函数中设置延迟的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！