从 QuickBlox 获取随机最终用户 [英] Getting random end user from QuickBlox

问题描述

我正在开发一款社交应用.对于当前用户，我需要随机显示该应用的其他用户.

I'm developing a social app. For current user, I need to randomly show other users of the app.

我首先想用这个代码来获取用户数:

I first want to use this code to get the number of users:

PagedRequest *request = [PagedRequest new];
request.perPage = 1;
request.page = 1;
[QBUsers usersWithPagedRequest:request delegate:self.proxy];

我注意到这是返回的 XML:

I noticed that this is the XML returned:

<users type="array" current_page="1" per_page="1" total_entries="13">
  <user>
    <blob-id type="integer" nil="true"/>
    <created-at type="datetime">2013-07-24T06:16:13Z</created-at>
    <email>[e-mail]</email>
    <external-user-id type="integer" nil="true"/>
    <facebook-id nil="true"/>
    <full-name>XXX xxx</full-name>
    <id type="integer">236286</id>
    <last-request-at type="datetime">2013-11-15T06:27:09Z</last-request-at>
    <login>t2wu</login>
    <owner-id type="integer">4282</owner-id>
    <phone nil="true"/>
    <twitter-id nil="true"/>
    <updated-at type="datetime">2013-11-15T06:27:09Z</updated-at>
    <website nil="true"/>
    <user-tags nil="true"/>
  </user>
</users>

我的意图是从total_entries=13"中得到13，所以我知道用户总数，然后再次发出这样的请求:

My intention is to get the 13 from the "total_entries=13", so I know that the total number of users, then make a request like this again:

PagedRequest *request = [PagedRequest new];
request.perPage = 1;
request.page = [self randomAmong: 13];
[QBUsers usersWithPagedRequest:request delegate:self.proxy];

where [self randomAmong: 13] 只需从 1 和 13 中选择一个数字，然后抓取用户的头像并显示给当前用户.

where [self randomAmong: 13] just pick one number out of 1 and 13, then grab the user's profile picture and show it to the current user.

三个问题:

1. 虽然我可以在控制台看到XML中的数字13输出，我不知道如何获得该数字.

1. Although I can see the number 13 in the XML in the console output, I have no idea how to get that number.

我只想轮流使用我的移动应用程序用户，而不是我的开发人员帐户(帐户用户).QuickBlox 允许多个应用程序与一个 QuickBlox 帐户用户相关联；然而，所有用户似乎被混为​​一谈.虽然我可以确定我只有此帐户内的一个应用程序，我仍然需要能够将自己排除在外.

I only want to rotate through my mobile application users, not my developer's account (the account user). QuickBlox allows multiple applications associated with one QuickBlox account user; however, all users seems to be lumped together. Although I could make sure that I only have one application inside this account, I still need to be able to exclude myself from it.

这是一个相对次要的问题.虽然上面获取当前用户数的尝试并没有尝试登录，委托方法-(void)completedWithResult:(Result*)result 将结果作为一个类QBUUserLogInResult.我不明白这是为什么，为什么不是QBUUserPagedResult 或 QBUUserResult 似乎使更多感觉?

This is a relatively minor question. Although the above attempt to get the current number of users is not an attempt to log in, the delegate method -(void)completedWithResult:(Result*)result has result as a class of QBUUserLogInResult. I don't understand why it is, why isn't it QBUUserPagedResult or QBUUserResult which seems to make more sense?

谢谢.

推荐答案

这是一个正确的方法，以下是您问题的答案:

This is a right way, here are the answers for your questions:

1)

#pragma mark - 
#pragma mark QBActionStatusDelegate

- (void)completedWithResult:(Result *)result{ 

    // Get User result 
    if(result.success && [result isKindOfClass:[QBUUserPagedResult class]]){
        QBUUserPagedResult *usersResult = (QBUUserPagedResult *)result; 

        int totalEntries = usersResult.totalEntries;
    }
}

2)

您可以使用标签来分隔用户.在用户注册时 - 只需为此用户添加一个标签，例如app1".

You can use tags to separate users. While user sign up - just add a tag to this user, for example 'app1'.

下一步 - 检索带有标签app1"的用户 - 使用下一个查询:

Next - to retrieve users with tag 'app1' - use the next query:

[QBUsers usersWithTags:[NSArray arrayWithObjects:@"app1", nil] delegate:self];

3)

用户登录查询将返回'QBUUserLogInResult'结果

User login query will return 'QBUUserLogInResult' result

获取单用户查询将返回'QBUUserResult'结果

Get single user query will return 'QBUUserResult' result

获取多个用户查询将返回'QBUUserPagedResult'结果

Get multiple users query will return 'QBUUserPagedResult' result

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