如何在Android onClick函数中设置延迟
) 我正在创建一个记忆游戏。我的问题是,每当我第二次单击时,我什至看不到切换按钮。需要明确的是 - 第一次单击会切换切换按钮,这样我就可以看到它所保存的数字,第二次单击不同的切换按钮应该会切换它,显示数字,然后继续设置分数+1(如果数字是)相同,或者如果不同则再次反转。
下面是我用作 onClick 函数的代码,我一直在考虑在第二个“if 块”中的某个位置放置某种睡眠或延迟函数 - (if(klikniecia ==2))。
任何有关此主题的帮助将不胜感激。
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;
}
}
}
}
}
)
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.
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)).
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;
}
}
}
}
}
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您可以将延迟消息发布到处理程序(带有关联的可运行程序)并让它更新 UI,而不是在 onclick 中休眠。显然,将其融入到应用程序的设计中并使其正常工作,但基本思想是这样的:
然后从 onClick 向处理程序发布一条延迟消息,指定要执行的可运行对象。
根据您的游戏的复杂程度,这可能并不完全是您想要的,但它应该给您一个开始。
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:
Then from onClick you post a delayed message to a handler, specifying that runnable to be executed.
Depending on how complicated your game is, this might not be exactly what you want, but it should give you a start.
正确工作:
Correctly work:
你永远不应该在 UI 线程中休眠(默认情况下,你的所有代码都在 UI 线程中运行)——它只会使 UI 冻结,不会让某些内容发生更改或完成。我无法提出更多建议,因为我不明白代码逻辑。
You never should sleep in UI thread (by default all your code runs in UI thread) - it will only make UI freeze, not let something change or finish. I can't suggest more because I don't understand code logic.
不要在 UI 线程中休眠。您需要另一个线程来查找来自 UI 线程的“唤醒并等待”消息。然后,第二个线程可以在正常的睡眠调用后进行隐藏。然后,您可以在下次需要隐藏某些内容时保留该线程,或者在每次需要另一次延迟时杀死它并启动一个新线程。
我相信有一些“postFoo”函数在这种情况下可能有用(从 UI 线程外部触发 UI 事件)。
Do not sleep in the UI thread. You need another thread that will look for a "wake up and wait" message from your UI thread. That second thread could then do your hiding after a normal sleep call. You could then keep the thread around for the next time you need to hide something, or kill it and whip up a new one each time you need another delay.
I believe there are some "postFoo" functions that might be useful in this context (triggering UI events from outside the UI thread).