Gan*_*hik 7 android xmpp smack asmack
我想阻止聊天列表中的某位朋友XMPP.代码工作正常.没有例外,但我无法阻止用户.我正在使用开火服务器.我应该在服务器上做些什么改变?
你们有什么想法吗?
我的代码:
public void XMPPAddNewPrivacyList(Connection connection, String userName) {
String listName = "newList";
// Create the list of PrivacyItem that will allow or
// deny some privacy aspect
List<PrivacyItem> privacyItems = new Vector<PrivacyItem>();
PrivacyItem item = new PrivacyItem(PrivacyItem.Type.jid.toString(),
false, 1);
item.setValue(userName);
privacyItems.add(item);
// Create the new list.
try {
PrivacyListManager privacyManager = new PrivacyListManager(connection);
privacyManager = PrivacyListManager
.getInstanceFor(connection);
privacyManager.createPrivacyList(listName, privacyItems);
} catch (XMPPException e) {
System.out.println("PRIVACY_ERROR: " + e);
}
}
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尝试这个 ...
public boolean blockFriend(String friendName) {
PrivacyItem item=new PrivacyItem(PrivacyItem.Type.jid,friendName, false, 7);
PrivacyListManager privacyManager = PrivacyListManager.getInstanceFor(connection);
List<PrivacyItem> list=new ArrayList<PrivacyItem>();
list.add(item);
try {
privacyManager.updatePrivacyList(NEWLIST, list);
privacyManager.setActiveListName(NEWLIST);
return true;
} catch (SmackException.NoResponseException |XMPPException.XMPPErrorException | SmackException.NotConnectedException e) {
e.printStackTrace();
return false;
}
}
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对于unblock,只需在privacyitem对象中将false替换为true即可。
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