如何修复此代码中的弃用警告?或者,还有其他的选择吗?
Handler().postDelayed({
context?.let {
//code
}
}, 3000)
如何修复此代码中的弃用警告?或者,还有其他的选择吗?
Handler().postDelayed({
context?.let {
//code
}
}, 3000)
当前回答
使用Executor而不是handler获取更多信息。 使用ScheduledExecutorService实现post delay:
ScheduledExecutorService worker = Executors.newSingleThreadScheduledExecutor();
Runnable runnable = () -> {
public void run() {
// Do something
}
};
worker.schedule(runnable, 2000, TimeUnit.MILLISECONDS);
其他回答
使用Executor而不是handler获取更多信息。 使用ScheduledExecutorService实现post delay:
ScheduledExecutorService worker = Executors.newSingleThreadScheduledExecutor();
Runnable runnable = () -> {
public void run() {
// Do something
}
};
worker.schedule(runnable, 2000, TimeUnit.MILLISECONDS);
根据文档(https://developer.android.com/reference/android/os/Handler#Handler()):
Implicitly choosing a Looper during Handler construction can lead to bugs where operations are silently lost (if the Handler is not expecting new tasks and quits), crashes (if a handler is sometimes created on a thread without a Looper active), or race conditions, where the thread a handler is associated with is not what the author anticipated. Instead, use an Executor or specify the Looper explicitly, using Looper#getMainLooper, {link android.view.View#getHandler}, or similar. If the implicit thread local behavior is required for compatibility, use new Handler(Looper.myLooper()) to make it clear to readers.
我们应该停止使用没有Looper的构造函数,而是指定一个Looper。
使用生命周期范围会更容易。内部活动或片段。
lifecycleScope.launch {
delay(2000)
// Do your stuff
}
或者使用处理器
Handler(Looper.myLooper()!!)
使用这个
Looper.myLooper()?.let {
Handler(it).postDelayed({
//Your Code
},2500)
}
只有无参数的构造函数已弃用,现在最好通过loop . getmainlooper()方法在构造函数中指定循环器。
用于Java
new Handler(Looper.getMainLooper()).postDelayed(new Runnable() {
@Override
public void run() {
// Your Code
}
}, 3000);
将它用于Kotlin
Handler(Looper.getMainLooper()).postDelayed({
// Your Code
}, 3000)
来源:developer.android.com