核心数据:NSPredicate用于多对多关系.("这里不允许使用多个键")

Question

我有两个名为"Category"和"Article"的实体,它们有很多关系.我想形成一个谓词,搜索category.name等于某个值的所有文章.我有以下内容:

 NSEntityDescription  *entityArticle   = [NSEntityDescription entityForName:@"Article" inManagedObjectContext:managedObjectContext]; 
 NSSortDescriptor  *sortDescriptor   = [[NSSortDescriptor alloc] initWithKey:@"title" ascending:YES];
 NSArray     *sortDescriptors  = [[NSArray alloc] initWithObjects:sortDescriptor, nil];
 NSPredicate    *predicate    = [NSPredicate predicateWithFormat:@"categories.name == [cd] %@", category.name]; 

 [request setSortDescriptors:sortDescriptors];
 [request setEntity:entityArticle];
 [request setPredicate:predicate];

 NSMutableArray *results = [[managedObjectContext executeFetchRequest:request error:nil] mutableCopy];

 if ([results count] > 0)
  NSLog(@"Results found."); 
 else 
  NSLog(@"NO results found."); 

 [request release];
 [sortDescriptor release];
 [sortDescriptors release];

我收到的错误是 *** Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: 'to-many key not allowed here'

有没有选择来检索所需的数据？

Answer 1

您正在尝试将collection(categories.name)与标量值(category.name)进行比较.您需要使用集合比较器(CONTAINS),或使用谓词修饰符(ANY/ ALL/ SOME等).

尝试使用:

[NSPredicate predicateWithFormat:@"ANY categories.name =[cd] %@", category.name];

要么:

[NSPredicate predicateWithFormat:@"categories.name CONTAINS[cd] %@", category.name];

Answer 2

快速语法

万一有人像我一样迅速地看到这篇文章......

let predicate = NSPredicate(format: "ANY categories.name = %@", category.name!)
fetchRequest.predicate = predicate

为我工作。