Question

Rewards -- Array
   Normal Reward -- Object
     Offer -- Dictionary
       Meta -- Dictionary
         rewardId -- String
   Normal Reward -- Object
     Offer -- Dictionary
       Meta -- Dictionary
         rewardId -- String

我有上面的层次结构，我必须为某个值匹配特定的 rewardId 。如何使用NSPredicate实现这一目标。

Answer 1

这个谓词应该可以解决问题：

NSPredicate *predicate = [NSPredicate predicateWithFormat:@"SELF.nameOfOfferProperty.nameOfMetaKey.nameOfRewardIdKey == %@", myRewardIdToMatch]

我根据您的需要明确命名了路径。

使用示例代码（这样您可以根据需要查看要使用的名称）：

NSMutableArray *rewards = [[NSMutableArray alloc] init];
for (NSUInteger i = 0; i < 10; i ++)
{
    NSString *rewardId = [NSString stringWithFormat:@"rewardID %@", (i%2 == 0)?@"Target":@"NonWanted"];
    NSString *otherMetaValue = [NSString stringWithFormat:@"otherMetaValue-%ld", i];
    NSString *otherOfferValue = [NSString stringWithFormat:@"otherOfferValue-%ld", i];
    NSDictionary *anOfferDict = @{@"meta": @{@"rewardId": rewardId,
                                             @"otherMetaKey": otherMetaValue},
                                  @"otherOfferKey": otherOfferValue};
    Reward *aReward = [[Reward alloc] initWithOfferDict:anOfferDict andIntValue:i];
    [rewards addObject:aReward];
}
NSLog(@"Rewards: %@", rewards);
NSString * myRewardIdToMatch = @"rewardID Target";
NSPredicate *predicate = [NSPredicate predicateWithFormat:@"SELF.offer.meta.rewardId == %@", myRewardIdToMatch];


NSArray *filteredRewards = [rewards filteredArrayUsingPredicate:predicate];
NSLog(@"FilteredRewards: %@", filteredRewards);

并且

@implementation Reward
-(id)initWithOfferDict:(NSDictionary *)dict andIntValue:(NSUInteger)intV
{
    self = [super init];
    if (self)
    {
        _offer = dict;
        _intV = intV;
    }
    return self;
}

覆盖description可能会有所帮助，并明确是否有效：

-(NSString *)description
{
    return [NSString stringWithFormat:@"<%@ %p> with IntValue: %ld and rewardID:\n%@", [self class], self, _intV,  _offer[@"meta"][@"rewardId"]];

}
@end

输出：

$> Rewards: (
    "<Reward 0x60000022ce80> with IntValue: 0 and rewardID: rewardID Target",
    "<Reward 0x60000022ce00> with IntValue: 1 and rewardID: rewardID NonWanted",
    "<Reward 0x60000022cf80> with IntValue: 2 and rewardID: rewardID Target",
    "<Reward 0x60000022cea0> with IntValue: 3 and rewardID: rewardID NonWanted",
    "<Reward 0x60000022cf20> with IntValue: 4 and rewardID: rewardID Target",
    "<Reward 0x60000022ce40> with IntValue: 5 and rewardID: rewardID NonWanted",
    "<Reward 0x60000022cfa0> with IntValue: 6 and rewardID: rewardID Target",
    "<Reward 0x60000022cfc0> with IntValue: 7 and rewardID: rewardID NonWanted",
    "<Reward 0x60000022ce60> with IntValue: 8 and rewardID: rewardID Target",
    "<Reward 0x60000022cec0> with IntValue: 9 and rewardID: rewardID NonWanted"
)
$> FilteredRewards: (
    "<Reward 0x60000022ce80> with IntValue: 0 and rewardID: rewardID Target",
    "<Reward 0x60000022cf80> with IntValue: 2 and rewardID: rewardID Target",
    "<Reward 0x60000022cf20> with IntValue: 4 and rewardID: rewardID Target",
    "<Reward 0x60000022cfa0> with IntValue: 6 and rewardID: rewardID Target",
    "<Reward 0x60000022ce60> with IntValue: 8 and rewardID: rewardID Target"
)

在iOS中使用词典中的字典数组的NSPredicate进行过滤

1 个答案: