Question

我保存到核心数据并使用下面的代码加载所有消息，但是如何仅为2个用户加载消息，对于自己的用户和目标用户，我希望将其基于用户JID。

NSFetchRequest *fetchRequest = [[NSFetchRequest alloc] init];

NSManagedObjectContext *context = [[self appDelegate].xmppMessageArchivingCoreDataStorage mainThreadManagedObjectContext];
NSEntityDescription *messageEntity = [NSEntityDescription entityForName:@"XMPPMessageArchiving_Message_CoreDataObject" inManagedObjectContext:context];
fetchRequest.entity = messageEntity;
NSSortDescriptor *sortDescriptor = [NSSortDescriptor sortDescriptorWithKey:@"timestamp" ascending:YES];
fetchRequest.sortDescriptors = [NSArray arrayWithObject:sortDescriptor];
NSError *error = nil;
NSArray *results = [context executeFetchRequest:fetchRequest error:&error];
//Now you get the NSArray with element type of "XMPPMessageArchiving_Message_CoreDataObject"

Answer 1

假设您的消息对象与from实体之间存在User关系就像添加（在执行请求之前）一样简单：

fetchRequest.predicate = [NSPredicate predicateWithFormat:@"from = %@ OR from = %@",selfUser,otherUser];

其中selfUser和otherUser是User核心数据管理对象或用户的托管对象ID。

如果您使用字段标识符（高度不推荐）：

 fetchRequest.predicate = [NSPredicate predicateWithFormat:@"bareJID.bare = %@ OR bareJID.bare = %@",selfUser.jid,otherUser.jid];

您可能需要阅读THIS

Answer 2

我使用了以下方法并且工作正常：）

-(void)checkForPastMessageAndDisplayIt
{
    NSString *strmyid = @"user1@domain";
    NSString *strBuddy = @"user2@domain";

    NSManagedObjectContext *moc = [[self appDelegate] managedObjectContext_message_achiving];
    NSEntityDescription *entity = [NSEntityDescription entityForName:@"XMPPMessageArchiving_Message_CoreDataObject"
                                              inManagedObjectContext:moc];

    NSFetchRequest *fetchRequest = [[NSFetchRequest alloc] init];
    fetchRequest.predicate = [NSPredicate predicateWithFormat:@"bareJidStr = %@ OR bareJidStr = %@", strmyid, strBuddy];
    [fetchRequest setEntity:entity];

    NSError *error;
    NSArray *messagesArray = [moc executeFetchRequest:fetchRequest error:&error];
    [messagesArray enumerateObjectsUsingBlock:^(id obj, NSUInteger idx, BOOL *stop) {
        id messageObject = [messagesArray objectAtIndex:idx];
        if ([messageObject isKindOfClass:[XMPPMessageArchiving_Message_CoreDataObject class]]) {

            XMPPMessageArchiving_Message_CoreDataObject *currentMessageCoreDataObject = (XMPPMessageArchiving_Message_CoreDataObject*)messageObject;
            XMPPMessage *message = currentMessageCoreDataObject.message;
            if (message!=nil) {
                [messages addObject:message];
            }

        }
    }];
}

Answer 3

你可以这样使用

-(void)testMessageArchiving {

    NSString *strmyid = @"myid";
    NSString *strBuddy ="toid";
    NSLog(@"my id and buddy is is %@ and %@",strmyid,strBuddy);

    XMPPMessageArchivingCoreDataStorage *storage = [XMPPMessageArchivingCoreDataStorage sharedInstance];
    NSManagedObjectContext *moc = [storage mainThreadManagedObjectContext];
    NSEntityDescription *entityDescription = [NSEntityDescription entityForName:@"XMPPMessageArchiving_Message_CoreDataObject"
                                                         inManagedObjectContext:moc];
    NSFetchRequest *request = [[NSFetchRequest alloc]init];

    request.predicate = [NSPredicate predicateWithFormat:@"bareJidStr = %@ OR bareJidStr = %@", strmyid, strBuddy];

    NSError *error;
    NSArray *messages = [moc executeFetchRequest:request error:&error];

    // NSLog(@"the history of messages is %@",messages);

    [self print:[[NSMutableArray alloc]initWithArray:messages]];

XMPP，CoreData - 如何仅加载自己和目标用户消息？

3 个答案: