如何使用XMPPFramework从ejabberd中删除两个用户之间的特定消息？

import XMPPFramework

// 登录到XMPP账号
let jid = XMPPJID(string: "username@domain.com")
let password = "password"
xmppStream.myJID = jid
xmppStream.connect(withTimeout: XMPPStreamTimeoutNone, error: nil)
xmppStream.authenticate(withPassword: password, error: nil)

// 获取消息历史记录
let xmppMessageArchivingStorage = XMPPMessageArchivingCoreDataStorage.sharedInstance()
let xmppMessageArchiving = XMPPMessageArchiving(messageArchivingStorage: xmppMessageArchivingStorage)
xmppMessageArchiving.activate(xmppStream)

// 检索特定消息
let userJID = XMPPJID(string: "user2@domain.com")
let messageContext = xmppMessageArchiving.messageArchivingManagedObjectContext()
let messageFetchRequest = NSFetchRequest<NSFetchRequestResult>(entityName: "XMPPMessageArchiving_Message_CoreDataObject")
messageFetchRequest.predicate = NSPredicate(format: "bareJidStr == %@ AND streamBareJidStr == %@", userJID.bare(), jid.bare())
let messages = try? messageContext.fetch(messageFetchRequest) as? [XMPPMessageArchiving_Message_CoreDataObject]

// 删除消息
if let messages = messages {
    for message in messages {
        messageContext.delete(message)
    }
    try? messageContext.save()
}