let iq1 = DDXMLElement(name: "iq") iq1.addAttribute(withname: "type",stringValue: "get") iq1.addAttribute(withname: "ID",stringValue: "0") let retrIEve = DDXMLElement(name: "retrIEve",xmlns: "urn:xmpp:archive") retrIEve?.addAttribute(withname: "with",stringValue: "[email protected]") let set = DDXMLElement(name: "set",xmlns: "http://jabber.org/protocol/rsm") let max = DDXMLElement(name: "max",stringValue: "50") iq1.addChild(retrIEve!) retrIEve?.addChild(set!) set?.addChild(max) stream?.send(iq1 )
现在我没有从双方得到历史.这是我得到的结果.
<iq xmlns="jabber:clIEnt" type="result" ID="0" to="[email protected]/5qsinh1syg"><chat xmlns="urn:xmpp:archive" with="[email protected]" start="2017-02-07T06:28:33.691Z"> <from secs="0"> <body>hello</body> </from> <from secs="405"> <body>hiii</body> </from> <from secs="580"> <body>hey</body> </from> <from secs="599"> <body>HooooE</body> </from> <from secs="628"> <body>hOo</body> </from> <from secs="907"> <body>Raman here</body> </from> <from secs="931"> <body>Raman here ONE</body> </from> <from secs="1273"> <body>Raman here</body> </from> <from secs="1558"> <body> Raman from SPark </body> </from> <from secs="1571"> <body> Raman from SPark One</body> </from> <from secs="1782"> <body>Raman from SPark </body> </from> <from secs="2651"> <body>Raman FiVE</body> </from> <from secs="2810"> <body>Raman Three</body> </from> <from secs="2810"> <body>Raman FOur</body> </from> <from secs="3534"> <body>RAMAN SIX</body> </from> <from secs="4105"> <body>Raman OKK</body> </from> <from secs="4113"> <body>OKKK l</body> </from> <from secs="4130"> <body>UST</body> </from> <set xmlns="http://jabber.org/protocol/rsm"> <first index="0">0</first> <last>17</last> <count>18</count> </set></chat></iq>
这是RAMAN不是BENTICK的所有消息,因为这两个用户参与聊天,同时获得历史,我只得到RAMAN的消息.不是BENTICk
如何从双方用户获取历史.
请帮忙.
解决方法 我已经实现了相同的方式来通过发送给定的iq来获取消息我得到了正确的结果.
let iQ = DDXMLElement.element(withname: "iq") as! DDXMLElement iQ.addAttribute(withname: "type",stringValue: "get") iQ.addAttribute(withname: "ID",stringValue: (xmppHandler.xmppStream?.generateUUID())!) let retrIEve = DDXMLElement(name: "retrIEve",xmlns: "urn:xmpp:archive") retrIEve?.addAttribute(withname: "with",stringValue: "\(JID!)@\(Utility.hostname)") let set = DDXMLElement.element(withname: "set") as! DDXMLElement set.addAttribute(withname: "xmlns",stringValue: "http://jabber.org/protocol/rsm") let max = DDXMLElement.element(withname: "max") as! DDXMLElement max.stringValue = "10" max.addAttribute(withname:"xmlns",stringValue: "http://jabber.org/protocol/rsm") iQ.addChild(retrIEve!) retrIEve?.addChild(set) set.addChild(max as DDXMLNode) xmppHandler.xmppStream?.send(iQ)<iq type="get" ID="7705EEC5-3385-42D4-B748-B39C204ADB90"><retrIEve xmlns="urn:xmpp:archive" with="[email protected]"><set xmlns="http://jabber.org/protocol/rsm"><max xmlns="http://jabber.org/protocol/rsm">10</max></set></retrIEve></iq>
响应
<iq xmlns="jabber:clIEnt" type="result" ID="7705EEC5-3385-42D4-B748-B39C204ADB90" to="[email protected]/Smack"> <chat xmlns="urn:xmpp:archive" with="[email protected]" start="2017-02-28T09:55:27.068Z"> <from secs="0" jID="[email protected]"> <body>Hiii rajeev</body> </from> <to secs="52"> <body>Hi How r u</body> </to> <from secs="15" jID="[email protected]"><body>I am good</body></from> <to secs="8"><body>What abt u</body></to> <to secs="169"><body>Hello testing chat</body></to> <from secs="10" jID="[email protected]"><body>Good</body></from> <from secs="5" jID="[email protected]"><body>Continue </body></from> <to secs="834"><body>Ok </body></to> <to secs="8"><body>Nice chat </body></to> <from secs="232" jID="[email protected]"><body>Heelo</body></from> <set xmlns="http://jabber.org/protocol/rsm"><first index="0">0</first> <last>9</last><count>574</count></set></chat></iq>总结
以上是内存溢出为你收集整理的ios – 在获取聊天记录时,我没有从Openfire获取用户历史记录全部内容，希望文章能够帮你解决ios – 在获取聊天记录时,我没有从Openfire获取用户历史记录所遇到的程序开发问题。
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