根据这个答案，我使用了setSoftInputMode方法，并覆盖了DialogFragment中的这些方法:

@Override
public void onCancel(@NonNull DialogInterface dialog) {
    super.onCancel(dialog);
    requireDialog().getWindow().setSoftInputMode(InputMethodManager.HIDE_IMPLICIT_ONLY);
}

@Override
public void onStart() {
    super.onStart();
    requireDialog().getWindow().setSoftInputMode(WindowManager.LayoutParams.SOFT_INPUT_STATE_ALWAYS_VISIBLE);
}

@Override
public void onStop() {
    super.onStop();
    requireDialog().getWindow().setSoftInputMode(InputMethodManager.HIDE_IMPLICIT_ONLY);
}

我也用这些方法创建了我自己的DialogFragment子类，所以当你创建另一个对话框并从这个类继承时，你会自动显示软键盘，而不需要任何其他编辑。希望对别人有用。

这很疯狂，但确实有效

fun showKeyboard(view: View) {
    try {
        InputMethodManager::class.java.getMethod(
                "showSoftInputUnchecked",
                Int::class.javaPrimitiveType,
                ResultReceiver::class.java
        ).apply {
            isAccessible = true
            invoke(view.context.inputMethodManager, 0, null)
        }
    }
    catch (e: Exception) {
       e.printStackTrace()
    }
}

我在各种不同的情况下都有同样的问题，我发现的解决方案在一些情况下工作，但在其他情况下不工作，所以这里是一个在我发现的大多数情况下工作的组合解决方案:

public static void showVirtualKeyboard(Context context, final View view) {
    if (context != null) {
        final InputMethodManager imm =  (InputMethodManager) context.getSystemService(Context.INPUT_METHOD_SERVICE);
        view.clearFocus();

        if(view.isShown()) {
            imm.showSoftInput(view, 0);
            view.requestFocus();
        } else {
            view.addOnAttachStateChangeListener(new View.OnAttachStateChangeListener() {
                @Override
                public void onViewAttachedToWindow(View v) {
                    view.post(new Runnable() {
                        @Override
                        public void run() {
                            view.requestFocus();
                            imm.showSoftInput(view, 0);
                        }
                    });

                    view.removeOnAttachStateChangeListener(this);
                }

                @Override
                public void onViewDetachedFromWindow(View v) {
                    view.removeOnAttachStateChangeListener(this);
                }
            });
        }
    }
}

以下是我从Square得到的一个更可靠的解决方案:

fun View.focusAndShowKeyboard() { /** * This is to be called when the window already has focus. */ fun View.showTheKeyboardNow() { if (isFocused) { post { // We still post the call, just in case we are being notified of the windows focus // but InputMethodManager didn't get properly setup yet. val imm = context.getSystemService(Context.INPUT_METHOD_SERVICE) as InputMethodManager imm.showSoftInput(this, InputMethodManager.SHOW_IMPLICIT) } } } requestFocus() if (hasWindowFocus()) { // No need to wait for the window to get focus. showTheKeyboardNow() } else { // We need to wait until the window gets focus. viewTreeObserver.addOnWindowFocusChangeListener( object : ViewTreeObserver.OnWindowFocusChangeListener { override fun onWindowFocusChanged(hasFocus: Boolean) { // This notification will arrive just before the InputMethodManager gets set up. if (hasFocus) { this@focusAndShowKeyboard.showTheKeyboardNow() // It’s very important to remove this listener once we are done. viewTreeObserver.removeOnWindowFocusChangeListener(this) } } }) } }

代码从这里开始。

没有一个答案对我有用。这里有一个简单的方法。

searchEditText.setVisibility(View.VISIBLE);
                final Handler handler=new Handler();
                handler.postDelayed(new Runnable() {
                    @Override
                    public void run() {
                        searchEditText.requestFocus();
                    }
                }, 400);

只是将requestFocus()方法延迟了400ms。

Kotlin扩展显示键盘的焦点。

这是之前的回答的组合，要么太长，要么不完整。

这个扩展在消息队列上发布了一个可运行的消息，在请求聚焦后显示软键盘:

fun View.showSoftKeyboard() {
    post {
        if (this.requestFocus()) {
            val imm = context.getSystemService(Context.INPUT_METHOD_SERVICE) as InputMethodManager
            imm?.showSoftInput(this, InputMethodManager.SHOW_IMPLICIT)
        }
    }
}

在需要时从任何视图调用它:

editText.showSoftKeyboard()