org.fourthline.cling.model.meta.Device.getIdentity()方法的使用及代码示例

x33g5p2x  于2022-01-18 转载在 其他  
字(6.1k)|赞(0)|评价(0)|浏览(137)

本文整理了Java中org.fourthline.cling.model.meta.Device.getIdentity()方法的一些代码示例，展示了Device.getIdentity()的具体用法。这些代码示例主要来源于Github/Stackoverflow/Maven等平台，是从一些精选项目中提取出来的代码，具有较强的参考意义，能在一定程度帮忙到你。Device.getIdentity()方法的具体详情如下：
包路径：org.fourthline.cling.model.meta.Device
类名称：Device
方法名：getIdentity

Device.getIdentity介绍

暂无

代码示例

代码示例来源：origin: 4thline/cling

  1. public DeviceItem(Device device) {
  2.   this.udn = device.getIdentity().getUdn();
  3.   this.device = device;
  4. }

代码示例来源：origin: 4thline/cling

  1. public DeviceItem(Device device, String... label) {
  2.   this.udn = device.getIdentity().getUdn();
  3.   this.device = device;
  4.   this.label = label;
  5. }

代码示例来源：origin: 4thline/cling

  1. protected String getDevicePath(Device device) {
  2.   if (device.getIdentity().getUdn() == null) {
  3.     throw new IllegalStateException("Can't generate local URI prefix without UDN");
  4.   }
  5.   StringBuilder s = new StringBuilder();
  6.   s.append(DEVICE).append("/");
  7.   s.append(URIUtil.encodePathSegment(device.getIdentity().getUdn().getIdentifierString()));
  8.   return s.toString();
  9. }

代码示例来源：origin: 4thline/cling

  1. @Override
  2.   public String toString() {
  3.     return "(" + getClass().getSimpleName() + ") Identity: " + getIdentity().toString() + ", Root: " + isRoot();
  4.   }
  5. }

代码示例来源：origin: 4thline/cling

  1. protected D find(UDN udn, D current) {
  2.   if (current.getIdentity() != null && current.getIdentity().getUdn() != null) {
  3.     if (current.getIdentity().getUdn().equals(udn)) return current;
  4.   }
  5.   if (current.hasEmbeddedDevices()) {
  6.     for (D embeddedDevice : (D[]) current.getEmbeddedDevices()) {
  7.       D match;
  8.       if ((match = find(udn, embeddedDevice)) != null) return match;
  9.     }
  10.   }
  11.   return null;
  12. }

代码示例来源：origin: 4thline/cling

  1. boolean contains(D device) {
  2.   return contains(device.getIdentity().getUdn());
  3. }

代码示例来源：origin: 4thline/cling

  1. protected Collection<D> findEmbeddedDevices(D current) {
  2.   Collection<D> devices = new HashSet<>();
  3.   if (!current.isRoot() && current.getIdentity().getUdn() != null)
  4.     devices.add(current);
  5.   if (current.hasEmbeddedDevices()) {
  6.     for (D embeddedDevice : (D[]) current.getEmbeddedDevices()) {
  7.       devices.addAll(findEmbeddedDevices(embeddedDevice));
  8.     }
  9.   }
  10.   return devices;
  11. }

代码示例来源：origin: 4thline/cling

  1. public ServiceReference getReference() {
  2.   return new ServiceReference(getDevice().getIdentity().getUdn(), getServiceId());
  3. }

代码示例来源：origin: 4thline/cling

  1. public List<ValidationError> validate() {
  2.   List<ValidationError> errors = new ArrayList<>();
  3.   if (getType() != null) {
  4.     // Only validate the graph if we have a device type - that means we validate only if there
  5.     // actually is a fully hydrated graph, not just a discovered device of which we haven't even
  6.     // retrieved the descriptor yet. This assumes that the descriptor will ALWAYS contain a device
  7.     // type. Now that is a risky assumption...
  8.     errors.addAll(getVersion().validate());
  9.     
  10.     if(getIdentity() != null) {
  11.       errors.addAll(getIdentity().validate());
  12.     }
  13.     if (getDetails() != null) {
  14.       errors.addAll(getDetails().validate());
  15.     }
  16.     if (hasServices()) {
  17.       for (Service service : getServices()) {
  18.         if (service != null)
  19.           errors.addAll(service.validate());
  20.       }
  21.     }
  22.     if (hasEmbeddedDevices()) {
  23.       for (Device embeddedDevice : getEmbeddedDevices()) {
  24.         if (embeddedDevice != null)
  25.           errors.addAll(embeddedDevice.validate());
  26.       }
  27.     }
  28.   }
  29.   return errors;
  30. }

代码示例来源：origin: 4thline/cling

  1. /**
  2.  * Returns root and embedded devices registered under the given UDN.
  3.  *
  4.  * @param udn A unique device name.
  5.  * @param rootOnly Set to true if only root devices (no embedded) should be searched
  6.  * @return Any registered root or embedded device under the given UDN, <tt>null</tt> if
  7.  *         no device with the given UDN has been registered.
  8.  */
  9. D get(UDN udn, boolean rootOnly) {
  10.   for (RegistryItem<UDN, D> item : deviceItems) {
  11.     D device = item.getItem();
  12.     if (device.getIdentity().getUdn().equals(udn)) {
  13.       return device;
  14.     }
  15.     if (!rootOnly) {
  16.       D foundDevice = (D)item.getItem().findDevice(udn);
  17.       if (foundDevice != null) return foundDevice;
  18.     }
  19.   }
  20.   return null;
  21. }

代码示例来源：origin: 4thline/cling

  1. private void establishLocalSubscription(LocalService service) {
  2.   if (getControlPoint().getRegistry().getLocalDevice(service.getDevice().getIdentity().getUdn(), false) == null) {
  3.     log.fine("Local device service is currently not registered, failing subscription immediately");
  4.     failed(null, null, new IllegalStateException("Local device is not registered"));

代码示例来源：origin: 4thline/cling

  1. incomingSubscription.getItem().getService().getDevice().getIdentity().getUdn();

代码示例来源：origin: kingthy/TVRemoteIME

  1. @Override
  2.   public String toString() {
  3.     return "(" + getClass().getSimpleName() + ") Identity: " + getIdentity().toString() + ", Root: " + isRoot();
  4.   }
  5. }

代码示例来源：origin: kingthy/TVRemoteIME

  1. boolean contains(D device) {
  2.   return contains(device.getIdentity().getUdn());
  3. }

代码示例来源：origin: 4thline/cling

  1. currentNode.add(deviceNode);
  2. addIfNotNull(deviceNode, "UDN: ", device.getIdentity().getUdn());
  3. addIfNotNull(deviceNode, "Device Type: ", device.getType().toString());
  4. if (device.getDetails().getDlnaDocs() != null) {

代码示例来源：origin: kingthy/TVRemoteIME

  1. public URI getPath(Device device) {
  2.   if (device.getIdentity().getUdn() == null) {
  3.     throw new IllegalStateException("Can't generate local URI prefix without UDN");
  4.   s.append(DEVICE).append("/");
  5.   s.append(URIUtil.encodePathSegment(device.getIdentity().getUdn().getIdentifierString()));

代码示例来源：origin: kingthy/TVRemoteIME

  1. public ServiceReference getReference() {
  2.   return new ServiceReference(getDevice().getIdentity().getUdn(), getServiceId());
  3. }

代码示例来源：origin: kingthy/TVRemoteIME

  1. protected D find(UDN udn, D current) {
  2.   if (current.getIdentity().getUdn().equals(udn)) return current;
  3.   if (current.hasEmbeddedDevices()) {
  4.     for (D embeddedDevice : (D[]) current.getEmbeddedDevices()) {
  5.       D match;
  6.       if ((match = find(udn, embeddedDevice)) != null) return match;
  7.     }
  8.   }
  9.   return null;
  10. }

代码示例来源：origin: kingthy/TVRemoteIME

  1. /**
  2.  * Returns root and embedded devices registered under the given UDN.
  3.  *
  4.  * @param udn A unique device name.
  5.  * @param rootOnly Set to true if only root devices (no embedded) should be searched
  6.  * @return Any registered root or embedded device under the given UDN, <tt>null</tt> if
  7.  *         no device with the given UDN has been registered.
  8.  */
  9. D get(UDN udn, boolean rootOnly) {
  10.   for (RegistryItem<UDN, D> item : deviceItems) {
  11.     D device = item.getItem();
  12.     if (device.getIdentity().getUdn().equals(udn)) {
  13.       return device;
  14.     }
  15.     if (!rootOnly) {
  16.       D foundDevice = (D)item.getItem().findDevice(udn);
  17.       if (foundDevice != null) return foundDevice;
  18.     }
  19.   }
  20.   return null;
  21. }

代码示例来源：origin: 4thline/cling

  1. deviceModelDetails.getSerialNumber()
  2. );
  3. appendNewElementIfNotNull(descriptor, deviceElement, ELEMENT.UDN, deviceModel.getIdentity().getUdn());
  4. appendNewElementIfNotNull(
  5.     descriptor, deviceElement, ELEMENT.presentationURL,

内容来源于网络，如有侵权，请联系作者删除！

Device

相关文章

    查看更多

    热门标签

    更多
    JavaquerypythonNode开发语言requestUtil数据库Table后端算法LoggerMessageElementParser

    最新文章

    更多
    • 蜀ICP备13028337号-1 大数据知识库 https://www.saoniuhuo.com © All rights reserved
    • 本站内容来源互联网,如果侵犯您的权益请联系我们删除, 联系方式：448109455@qq.com