我最近尝试使用读取传入通知AccessibilityService.我知道阅读通知使用NotificationListenerService,但这不是我需要的(为了兼容较低的Android版本).我的问题是,OnServiceConnected()永远不会被调用,即使在我的设置中,我已经为我的应用程序提供了必要的预测.这是我的代码:
public class NotificationAccessibilityService extends AccessibilityService{
protected void onServiceConnected() {
Log.d("Tortuga", "AccessibilityService Connected");
AccessibilityServiceInfo info = new AccessibilityServiceInfo();
info.eventTypes = AccessibilityEvent.TYPE_NOTIFICATION_STATE_CHANGED;
info.feedbackType = AccessibilityServiceInfo.FEEDBACK_ALL_MASK;
info.notificationTimeout = 100;
setServiceInfo(info);
}
@Override
public void onAccessibilityEvent(AccessibilityEvent e) {
Log.d("Tortuga","FML");
if (e.getEventType() == AccessibilityEvent.TYPE_NOTIFICATION_STATE_CHANGED) {
Log.d("Tortuga","Recieved event");
Parcelable data = e.getParcelableData();
if (data instanceof Notification) {
Log.d("Tortuga","Recieved notification");
Notification notification = (Notification) data;
Log.d("Tortuga","ticker: " + notification.tickerText);
Log.d("Tortuga","icon: " + notification.icon);
Log.d("Tortuga", "notification: "+ e.getText());
}
}
} …Run Code Online (Sandbox Code Playgroud) xml notifications android accessibility accessibilityservice
我是旋转模型检查的新手,想知道这个错误意味着什么:
unreached in proctype P1
ex2.pml:16, state 11, "-end-"
(1 of 11 states)
unreached in proctype P2
ex2.pml:29, state 11, "-end-"
(1 of 11 states)
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这是我的代码:
int y1, y2;
byte insideCritical;
active proctype P1(){
do
::true->
y2 = y1 + 1;
(y1 == 0 || y2 < y1);
/* Entering critical section */
insideCritical++;
assert(insideCritical < 2);
insideCritical--;
/* Exiting critical section */
y2 = 0;
od
}
active proctype P2(){
do
::true->
y1 = y2 + 1;
(y2 == …Run Code Online (Sandbox Code Playgroud) 我的aspx中有一个转发器:
<asp:Repeater ID="rptDummy" runat="server" OnItemDataBound="rptDummy_OnItemDataBound"
Visible="true">
</asp:Repeater>
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在网络的c#侧我写了这个函数:
protected void createRadioButtons(DataSet ds){
List<System.Web.UI.WebControls.RadioButton> buttons = new List<System.Web.UI.WebControls.RadioButton>();
foreach (DataTable dt in ds.Tables){
foreach (DataRow r in dt.Rows){
System.Web.UI.WebControls.RadioButton rb = new System.Web.UI.WebControls.RadioButton();
rb.Text = r[1] + " " + r[2] + " " + r[3] + " " + r[4];
rb.GroupName = (string)r[5];
buttons.Add(rb);
}
}
rptDummy.DataSource = buttons;
rptDummy.DataBind();
}
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但是在尝试时,它什么都没有显示出来.我究竟做错了什么?
有没有办法从python启动和停止进程?我说的是在正常运行时用 ctrl+z 停止的继续进程。我想启动这个过程,等待一段时间然后终止它。我正在使用Linux。
这个问题不像我的,因为在那里,用户只需要运行这个过程。我需要运行它并停止它。
我正在构建一个应用程序,它会要求用户将聊天从 WhatsApp 导出到我的应用程序中。如何在“通过...发送聊天”意图窗口中显示我的应用程序?
我在我的项目中使用 ROS,我需要不时发送一条消息。我有这个功能:
void RosNetwork::sendMessage(string msg, string channel) {
_mtx.lock();
ros::Publisher chatter_pub = _n.advertise<std_msgs::String>(channel.c_str(),10);
ros::Rate loop_rate(10);
std_msgs::String msgToSend;
msgToSend.data = msg.c_str();
chatter_pub.publish(msgToSend);
loop_rate.sleep();
cout << "Message Sent" << endl;
_mtx.unlock();
}
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我在python中有这个:
def callbackFirst(data):
#rospy.loginfo(rospy.get_caller_id() + "I heard %s", data.data)
print("Received message from first filter")
def callbackSecond(data):
#rospy.loginfo(rospy.get_caller_id() + "I heard %s", data.data)
print("Received message from second filter")
def listener():
rospy.Subscriber("FirstTaskFilter", String, callbackFirst)
print("subscribed to FirstTaskFilter")
rospy.Subscriber("SecondTaskFilter", String, callbackSecond)
print("subscribed to SecondTaskFilter")
rospy.spin()
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侦听器是python中的一个线程。我进入了该函数sendMessage(我在终端中看到“Message Sent”很多次),但我没有看到 python 脚本接收到消息。
更新:我测试了 python …