核心数据连接表，有许多通过，获取属性谓词 [英] Core data join table, has many through, fetched properties predicate

问题描述

如何为将要呈现的地方实体编写提取的财产一个用户数组？

How do you write a Fetched Property for the Place Entity that will present an Array of Users?

推荐答案

对于提取的属性用户的 Place 检索所有签到与指定地点相关的用户，设置

For a fetched property users of Place that retrieves all users whose check-in is related to the given place, set

• 获取的目的地属性用户


• 和谓词ANY checkins.event == $ FETCH_SOURCE

现在你可以获得一个地方的用户数组：

Now you can get the array of users for a place:

Place *place = ...;
NSArray *users = [place valueForKey:@"users"];

此获取的属性对应于以下获取请求：

This fetched property corresponds to the following fetch request:

Place *place = ...;
NSFetchRequest *request = [NSFetchRequest fetchRequestWithEntityName:@"User"];
NSPredicate *predicate = [NSPredicate predicateWithFormat:@"ANY checkins.event == %@", place];
[request setPredicate:predicate];
NSArray *users = [context executeFetchRequest:request error:&error];

如果您将获取的属性 users 声明为动态属性：

If you declare the fetched property users as a dynamic property:

@interface Place (FetchedProperties)
@property(nonatomic, retain) NSArray *users;
@end

@implementation Place (FetchedProperties)
@dynamic users;
@end

然后您可以使用属性语法检索值：

then you can retrieve the value using property syntax:

NSArray *users = place.users;
// instead of: NSArray *users = [place valueForKey:@"users"];

但请注意 您可以获得直接相同的结果（作为一组），不使用fetched属性：

But note that you can get the same result (as a set) directly, without using a fetched property:

Place *place = ...;
NSSet *users = [place.checkins valueForKey:@"user"];

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