【发布时间】:2011-09-28 06:12:13
【问题描述】:
我有一个具有一对多关系的核心数据模式,但我正在努力获得价值,例如
Features.h 模型如下:
@interface Features : NSManagedObject
@property (nonatomic, retain) NSString * name;
@property (nonatomic, retain) NSString * version;
@property (nonatomic, retain) NSNumber * expires;
@property (nonatomic, retain) NSNumber * used;
@property (nonatomic, retain) NSNumber * licenses;
@property (nonatomic, retain) NSManagedObject *featureToProducts;
@property (nonatomic, retain) NSSet *featureToUsers;
@property (nonatomic, retain) NSManagedObject *featureToDaemons;
@end
@interface Features (CoreDataGeneratedAccessors)
- (void)addFeatureToUsersObject:(NSManagedObject *)value;
- (void)removeFeatureToUsersObject:(NSManagedObject *)value;
- (void)addFeatureToUsers:(NSSet *)values;
- (void)removeFeatureToUsers:(NSSet *)values;
@end
并尝试像这样访问它:
Features *feature = (Features *)[[self.sectionInfoArray objectAtIndex:indexPath.section] features];
NSLog(@"featureToUsers = %@", [feature.featureToUsers valueForKey:@"user"]);
NSLog 显示:
2011-09-28 01:06:59.719 产品名称[74092:fb03] featureToUsers = {( “大卫.威尔逊”)}
我正在尝试填充 UITableViewCell 但我不断收到:
'NSInvalidArgumentException',原因:'-[__NSCFSet isEqualToString:]: 无法识别的选择器发送到实例 0x6b59a50'
谢谢, 狐狸
【问题讨论】:
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抱歉,我正在尝试将 [feature.featureToUsers valueForKey:@"user"] 作为字符串。例如“david.wilson”,但我不断收到:'NSInvalidArgumentException',原因:'-[__NSCFSet isEqualToString:]: unrecognized selector sent to instance 0x6b59a50'
标签: cocoa core-data uitableview