Question

我无法从包含许多NSArray的{{1}}创建索引，我只想根据字典中的NSDictionaries键来索引值。例如，每个字典看起来像这样：

username

目前我已简化了问题，因此我正在为{ username => "daspianist", //<- Only want to use this value to create index objectId => "hjd72h3jd", createdAt => "30-1-2014", updatedAt => "30-1-2014" } NSArray编制索引，我的工作如下：

NSStrings

我正在努力将我为NSStrings所做的事情转换为我的NSDictionaries中//Note that `stringArray` is passed to this method NSMutableDictionary *dict = [NSMutableDictionary dictionary]; for (char firstChar = 'a'; firstChar <= 'z'; firstChar++) { NSString *firstCharacter = [NSString stringWithFormat:@"%c", firstChar]; NSArray *content = [stringArray filteredArrayUsingPredicate:[NSPredicate predicateWithFormat:@"SELF beginswith[cd] %@", firstCharacter]]; NSMutableArray *mutableContent = [NSMutableArray arrayWithArray:content]; if ([mutableContent count] > 0) { NSString *key = [firstCharacter uppercaseString]; [dict setObject:mutableContent forKey:key]; NSLog(@"%@: %u", key, [mutableContent count]); } }键的值，并且不胜感激。谢谢！

更新

为了澄清，我感兴趣的结果词典看起来像这样

username

更新2

根据{ "a" => { username => "aardvark", otherKeys => "otherValues" }, { username => "applepicking", otherKeys => "otherValues" } "d" => { username => "daspianist", otherKeys => "otherValues" } .... }的答案输入解决方案，以方便使用：

Wain

Answer 1

尝试实现以下功能：

- (NSArray *)sectionIndexTitlesForTableView:(UITableView *)tableView
{
    NSMutableArray *charactersForSort = [[NSMutableArray alloc] init];
    for (NSDictionary *item in array)
    {
        NSString *username = [item objectForKey:@"username"]
        if (![charactersForSort containsObject:[username substringToIndex:1]])
        {
            [charactersForSort addObject:[username substringToIndex:1]];
        }
    }

    return charactersForSort;
}

此处数组是一个 NSArray 的词典，您已在问题中提及。

Answer 2

因此，content是与当前密钥匹配的username数组。目标是找到该数组中包含username的所有词典。这是谓词的工作：

NSPredicate *p = [NSPredicate predicateWithFormat:@"username IN %@", content];

现在你可以这样做：

NSArray *values = [dictArray filteredArrayUsingPredicate:p];
[dict setObject:values forKey:key];

根据NSDiction中NSArray内的值为UITableView创建索引

2 个答案: