给定 JSON 对象的 NSArray，如何在 NSDictionary 中存储同一个键的多个值？

Question

我知道不可能为同一个键设置多个值，除非这些值存储在数组中。

我有这个名为 friMainDicArray 的 JSON 对象的 NSArray：

{
        day = 0;
        "end_time" = "17:30";
        "english_event" = "Lion Dance";
        "english_performer" = "Legendary Group";
        "image_link" = "schedule_miss_vn";
        stage = 0;
        "start_time" = "17:00";
        "viet_event" = "<null>";
        "viet_performer" = "Nh?m Legendary";
    },
        {
        day = 0;
        "end_time" = "18:00";
        "english_event" = Singing;
        "english_performer" = "Ivan Cheong";
        "image_link" = "schedule_miss_vn";
        stage = 0;
        "start_time" = "17:30";
        "viet_event" = "Ca Nh?c";
        "viet_performer" = "Ivan Cheong";
    },
        {
        day = 0;
        "end_time" = "22:00";
        "english_event" = Singing;
        "english_performer" = "Between California and Summer";
        "image_link" = "";
        stage = 0;
        "start_time" = "21:00";
        "viet_event" = "Ca Nh?c";
        "viet_performer" = "";
    },
        {
        day = 0;
        "end_time" = "";
        "english_event" = "End of Day";
        "english_performer" = "";
        "image_link" = "";
        stage = 0;
        "start_time" = "22:00";
        "viet_event" = "";
        "viet_performer" = "";
    },
        {
        day = 0;
        "end_time" = "21:00";
        "english_event" = Music;
        "english_performer" = "DJ Happee From Channel 93.3";
        "image_link" = "";
        stage = 0;
        "start_time" = "20:00";
        "viet_event" = "";
        "viet_performer" = "";
    },
        {
        day = 0;
        "end_time" = "21:00";
        "english_event" = Music;
        "english_performer" = "Adam Cease";
        "image_link" = "";
        stage = 0;
        "start_time" = "20:00";
        "viet_event" = "";
        "viet_performer" = "";
    },
        {
        day = 0;
        "end_time" = "21:00";
        "english_event" = "Ao Dai Fashion Show";
        "english_performer" = "";
        "image_link" = "";
        stage = 0;
        "start_time" = "20:00";
        "viet_event" = "";
        "viet_performer" = "";
    }
}

我正在存储每个 JSON 对象，给定 start_time 键值，但问题在于 20:00 键。我想将 20:00 存储为键，将 JSON 对象存储为值。问题是，最后 3 个 JSON 对象包含相同的 20:00 键，所以我想将这 3 个 JSON 对象存储为 NSArray，并将其设置为 20:00 键。

我想实现这样的目标：

for (int i = 0; i < [friMainDicArray count]; i++)
{
    [friMainDic setValue:friMainDicArray[i] forKey:[friMainDicArray[i] valueForKey:@"start_time" ] ]
}

但是，我不知道检查start_time 键是否相同的逻辑，然后将其值添加到同一个键中。

有人能指点我吗？

Answer 1

我建议您将 JSON 对象始终存储在数组中，无论 keys 是否相同。通过这种方式，您至少可以避免在每次迭代时进行某种模糊检查，例如：

if ([dict[key] isKindOfClass:[NSMutableArray class]]) {
    NSMutableArray *values = (NSMutableArray *)dict[key];
    [values addObject:newObject];
}
else {
    NSMutableArray *values = [NSMutableArray arrayWithObject:dict[key]];
    [values addObject:newObject];
}

相反，我会使用以下方法通过遍历数组来填充结果字典：（假设所有 JSON 对象现在都是 NSDictionary 实例）

NSArray *events = @[@{@"start_time": @"17:00", @"end_time": @"17:30", @"english_event": @"Lion Dance", @"english_performer": @"Legendary Group"},
                    @{@"start_time": @"17:30", @"end_time": @"18:00", @"english_event": @"Singing", @"english_performer": @"Ivan Cheong"},
                    @{@"start_time": @"20:00", @"end_time": @"21:00", @"english_event": @"Music", @"english_performer": @"Adam Cease"},
                    @{@"start_time": @"20:00", @"end_time": @"21:00", @"english_event": @"Music", @"english_performer": @"DJ Happee From Channel 93.3"},
                    @{@"start_time": @"21:00", @"end_time": @"22:00", @"english_event": @"Singing", @"english_performer": @"Between California and Summer"}];

NSMutableDictionary *resultDict = [NSMutableDictionary dictionary];

[events enumerateObjectsUsingBlock:^(NSDictionary *event, NSUInteger idx, BOOL *stop) {

    NSMutableArray *values = resultDict[event[@"start_time"]];

    if (values == nil) {
        values = [[NSMutableArray alloc] init];
        resultDict[event[@"start_time"]] = values;
    }
    [values addObject:event];
}];

注意：我找不到更短的方法，包括collection operators、KVO accessors 等，实际上将使用0(n) 对字典进行分组。

Answer 2

好吧，我想我会这样处理它。找出是否已经有一个数组。制作一个可变版本并添加到它。把它放回字典里。如果没有创建一个数组并添加到它。

for (int i = 0; i < [friMainDicArray count]; i++)
{
    if (firMainDic[[friMainDicArray[i] valueForKey:@"start_time" ])
    {
        //pull out the array and add to it
        NSMutableArray *mutableArray = [[NSMutableArray alloc]initWithArray: firMainDic[[friMainDicArray[i] valueForKey:@"start_time" ]]];
        [mutableArray addObject:friMainDicArray[i]];
        [friMainDic setValue:mutableArray forKey:[friMainDicArray[i] valueForKey:@"start_time" ] ];
    }
    else
    {
        NSArray *array = @[friMainDicArray[i]];
        [friMainDic setValue:array forKey:[friMainDicArray[i] valueForKey:@"start_time" ] ]
    }
}

不是 100% 确定，但我相信当您将可变数组放入字典时，它会作为非可变数组出现。无论哪种方式，这至少应该让您朝着正确的方向开始。

Answer 3

我认为你可以循环数组，在 setValueForKey 之前，检查键是否已经存在，如果是，如果该键的值是 NSDictionary 类型，则将当前键与现有键组合为一个数组，否则该键的值应为 NSArray，然后将当前项添加到该数组中。

for (NSDictionary *item in friMainDicArray)
{
    NSString *key = [item valueForKey:@"start_time"];
    if ([friMainDic objectForKey:key]) {
        id value = [friMainDic objectForKey:key];
        if ([value isKindOfClass:[NSArray class]]) {
            NSMutableArray *array = [NSMutableArray arrayWithArray:(NSArray*)value];
            [array addObject:item];
            [friMainDic setValue:array forKey:key];
        } else {
            NSDictionary *dict = (NSDictionary*)value;
            NSArray *array = @[dict,item];
            [friMainDic setValue:array forKey:key];
        }
    } else {
        [friMainDic setValue:item forKey:key];
    }
}