我试图闪烁文本(基于之前的按钮点击(,但我的应用程序在1000ms/2000ms后崩溃。
我试过创建一个线程,但我不确定这是否是最简单的方法。有人能帮我吗?
这是我当前的代码:
fun toggleBlinkCounterUI() {
/*var handler: Handler = Handler();
Thread(Runnable() {
override fun run() {
var timeToBlink: Long = 1000;
Thread.sleep(timeToBlink)
handler.post(Runnable() {
fun run() {
if (binding.exerciseTimer.visibility == View.VISIBLE){
binding.exerciseTimer.visibility = View.INVISIBLE
} else {
binding.exerciseTimer.visibility = View.VISIBLE
}
toggleBlinkCounterUI();
}
});
}
}).start();*/
/*val thread = Thread {
var timeToBlink: Long = 1000;
while (1 == 1) {
Thread.sleep(timeToBlink)
if (binding.exerciseTimer.visibility == View.VISIBLE){
binding.exerciseTimer.visibility = View.INVISIBLE
} else {
binding.exerciseTimer.visibility = View.VISIBLE
}
}
}
thread.start()*/
提前感谢
使用Handler并利用其postDelayed方法。当您想要更改UI时,您需要使用主循环器:val uiCallbackHandler: Handler = Handler(Looper.getMainLooper())
。除此之外,你还需要两样东西:
- 创建可运行程序,它将定义UI交替逻辑
- 延迟后的1000毫秒,你目前使用线程睡眠
利用我的建议和你的代码的组合(我也让它更像Kotlin(:
val handler = Handler(Looper.getMainLooper())
val switchVisibilityRunnable = Runnable {
binding.exerciseTimer.isVisible = !binding.exerciseTimer.isVisible
handler.postDelayed(switchVisibilityRunnable, 1000)
}
handler.post(switchVisibilityRunnable)