键值编码获取列表中的元素

Question

我有两个对象：

@interface AObject : NSObject

@property NSArray *bObjects;

@end

@interface BObject : NSObject

@property NSString *name;

@end

在AObject的实例上使用键值编码，我可以获得bObjects的列表（ @"self.bObjects" ）和bObjects名称的列表（ @"self.bObjects.name" ）。

但是，我想要的只是第一个bObjects的名称。 我的直觉是键值编码应支持列表下标，例如： @"bObjects[0].name" 。

但这似乎并不存在。 我如何获得一个实体？ 使用键值编码的AObject的第一个BObject的名称？

脚注：我意识到在最后一个问题中，我愚蠢地将NSPredicate和KV编码混为一谈。

Answer 1

正如Martin R在评论中提到的那样，当前最好的选择是在AObject类中创建firstBObject属性。

AObject.h /米

@class BObject;

@interface AObject : NSObject
+ (AObject*)aObjectWithBObjects:(NSArray*)bObjects;
@property NSArray *bObjects;
@property (nonatomic, readonly) BObject *firstBObject;
@end

@implementation AObject
+ (AObject*)aObjectWithBObjects:(NSArray*)bObjects
{
    AObject *ao = [[self alloc] init];
    ao.bObjects = bObjects;
    return ao;
}
- (BObject*)firstBObject
{
    return [self.bObjects count] > 0 ? [self.bObjects objectAtIndex:0] : nil;
}
@end

BObject.h /米

@interface BObject : NSObject
+ (BObject*)bObjectWithName:(NSString*)name;
@property NSString *name;
@end

@implementation BObject
+ (BObject*)bObjectWithName:(NSString *)name
{
    BObject *bo = [[self alloc] init];
    bo.name = name;
    return bo;
}
@end

用法：

NSArray *aobjects = @[
                      [AObject aObjectWithBObjects:@[
                       [BObject bObjectWithName:@"A1B1"],
                       [BObject bObjectWithName:@"A1B2"],
                       [BObject bObjectWithName:@"A1B3"],
                       [BObject bObjectWithName:@"A1B4"]
                       ]],
                      [AObject aObjectWithBObjects:@[
                       [BObject bObjectWithName:@"A2B1"],
                       [BObject bObjectWithName:@"A2B2"],
                       [BObject bObjectWithName:@"A2B3"],
                       [BObject bObjectWithName:@"A2B4"]
                       ]],
                      [AObject aObjectWithBObjects:@[
                       [BObject bObjectWithName:@"A3B1"],
                       [BObject bObjectWithName:@"A3B2"],
                       [BObject bObjectWithName:@"A3B3"],
                       [BObject bObjectWithName:@"A3B4"]
                       ]]
                      ];
NSLog(@"%@", [aobjects valueForKeyPath:@"firstBObject.name"]);

结果

（
A1B1，
A2B1，
A3B1
）

Answer 2

事实证明，我很幸运能够简单地在根类（ AObject ）中重写-valueForKey: AObject 值得重申的是-valueForKeyPath:呼吁-valueForKey:在每一个关键，这是很酷。

由于这可能并不适用于所有人，并且可能过多地操纵了默认的预期行为，所以这绝对不是 “正确”的答案。

但是无论如何，这里是：

- (id)valueForKey:(NSString *)string
{
    if ([string characterAtIndex: [string length] - 1] == ']') // Trying to subscript
    {
        NSRegularExpression *subscriptRegex = [[NSRegularExpression alloc] initWithPattern: @"([a-zA-Z]+)\\[([0-9]+)\\]"
                                                                                   options: (NSRegularExpressionOptions)0
                                                                                     error: nil];

        NSString *key = [subscriptRegex stringByReplacingMatchesInString: string
                                                                 options: (NSMatchingOptions)0
                                                                   range: NSMakeRange(0, [string length])
                                                            withTemplate: @"$1"];
        id valueForKey = [self valueForKey: key];
        if (!key || !valueForKey || ![valueForKey respondsToSelector: @selector(objectAtIndexedSubscript:)])
            return nil;

        NSInteger index = [[subscriptRegex stringByReplacingMatchesInString: string
                                                                    options: (NSMatchingOptions)0
                                                                      range: NSMakeRange(0, [string length])
                                                               withTemplate: @"$2"] integerValue];
        if ((index < 0) || (index >= [valueForKey count]))
            return nil;

        return [valueForKey objectAtIndexedSubscript: index];
    }

    return [super valueForKey: string];
}