我有两张表TrendingUsers
和Follow
。所需的功能就像从TrendingUsers
表中获取用户并提供跟随,前提是获取的用户不是来自用户的关注列表。如果用户已经被跟踪,则跳过。
Follow
表格包含follower
和leader
列。
PFQuery *followTableQuery = [PFQuery queryWithClassName:@"Follow"];
[followTableQuery whereKey:@"follower" equalTo:[PFUser currentUser] ];
[followTableQuery whereKey:@"leader" equalTo:@"fetchedUserObject" ];
[followTableQuery findObjectsInBackgroundWithBlock:^(NSArray *objects, NSError *error) {
if (!error) {
if (objects.count) {
//if following objects array will have single object
}
else
{
//not following to @"fetchedUserObject" user
}
}
}
];
这将确认currentUser
是否跟随@"fetchedUserObject"
用户。
现在我想将此集成到TrendingUsers
表查询中,以仅获取currentUser
未跟踪的用户。
答案 0 :(得分:2)
您可以简单地使用嵌套查询,来自Parse的文档通常是一个很好的起点。这是一个示例代码,根据我从您的问题中理解,这应该可以解决问题。
//This is our current user
PFUser *user = [PFUser currentUser];
//The first query, querying for all the follow objects from the current user
PFQuery *followingQuery = [PFQuery queryWithClassName:@"Follow"];
[followingQuery whereKey:@"follower" equalTo:user];
//Now we query for the actual trending users, but we do not want the query to return the users (who are in the @"leader" key) that have been found by the first query
PFQuery *trendingQuery = [PFQuery queryWithClassName:@"TrendingUsers"];
[trendingQuery whereKey:@"objectId" notEqualTo:user.objectId]; //don't return the current user
[trendingQuery whereKey:@"objectId" doesNotMatchKey:@"leader" inQuery:followingQuery]; //I'm supposing that @"leader" is containing the objectId of the specific user that is part of the follow object with the current user
[trendingQuery setLimit:1000];
[trendingQuery findObjectsInBackgroundWithBlock:^(NSArray *objects, NSError *error)
{
//...
}];
我可能没有完全理解您的数据结构,因此您可能需要在上面的代码中交换一个或多个键,但基本上,这就是您要这样做的方法。