Smack API 抛出 item-not-found(404) 异常
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【中文标题】Smack API 抛出 item-not-found(404) 异常【英文标题】:Smack API throws item-not-found(404) exception 【发布时间】:2016-05-31 18:40:06 【问题描述】:这里是代码
Iterator i = MultiUserChat.getJoinedRooms(connectionManager.getXMPPConnection(), "test123@dulanjaya-pc");
while(i.hasNext())
System.out.println(i.next());
【问题讨论】:
发现问题;该代码不包含main(String[])
方法。此外,connectionManager
永远不会被初始化,并且它们都不包含在类中。
好吧。 connectionManager 已初始化。代码中没有提到
【参考方案1】:
想通了。 需要把服务作为/Smack
Iterator i = MultiUserChat.getJoinedRooms(connectionManager.getXMPPConnection(), "test123@dulanjaya-pc/Smack");
【讨论】:
【参考方案2】:也许这段代码 sn-p 对你有帮助。
boolean supports = MultiUserChat.isServiceEnabled(connection,"test3@pc2010102716/spark");
if(supports)
Iterator<String> joinedRooms = MultiUserChat.getJoinedRooms(connection,"test3@pc2010102716/spark");
while(joinedRooms.hasNext())
System.out.println("test3 has joined Room " + joinedRooms.next());
【讨论】:
【参考方案3】:getJoinedRooms 的源代码如下所示,告诉我们该方法的第二个参数需要一个完全限定的 xmpp ID。它显示的 ID 示例不完整,完整的 ID 应包含资源名称,例如 /smack、/spark 和 /android。
/**
* Returns an Iterator on the rooms where the requested user has joined. The Iterator will
* contain Strings where each String represents a room (e.g. room@muc.jabber.org).
*
* @param connection the connection to use to perform the service discovery.
* @param user the user to check. A fully qualified xmpp ID, e.g. jdoe@example.com.
* @return an Iterator on the rooms where the requested user has joined.
*/
public static Iterator<String> getJoinedRooms(Connection connection, String user)
try
ArrayList<String> answer = new ArrayList<String>();
// Send the disco packet to the user
DiscoverItems result =
ServiceDiscoveryManager.getInstanceFor(connection).discoverItems(user, discoNode);
// Collect the entityID for each returned item
for (Iterator<DiscoverItems.Item> items=result.getItems(); items.hasNext();)
answer.add(items.next().getEntityID());
return answer.iterator();
catch (XMPPException e)
e.printStackTrace();
// Return an iterator on an empty collection
return new ArrayList<String>().iterator();
【讨论】:
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