给定JSON对象的NSArray,如何在NSDictionary中为同一个键存储多个值?

我认为您可以循环数组,在setValueForKey之前,检查该键是否已经存在,如果是,如果该键的值是NSDictionary类型,则将当前值与现有值组合为数组,否则该键的应为NSArray,然后将当前项添加到该数组。

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];
    }
}

我想我会这样做。查找是否已经存在数组。创建一个可变版本并添加到其中。将其重新设置到字典中。如果没有,创建一个数组并添加到其中。

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,但我相信当您将一个可变数组放入字典时,它会变成一个非可变数组。不管怎样,这至少会让你朝着正确的方向开始。

我建议你把你的 JSON 对象始终在数组中,无论是否 keys 相同或不同。通过这种方式,您至少可以避免在每次迭代时进行一些模糊的检查,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,我将通过使用以下方法在数组上迭代一次来填充生成的字典:( 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: 我找不到更短的方法,包括 collection operators , KVO accessors 等等,这实际上需要 0(n) 用于对字典进行分组。