NSFetchedResultsController與索引UITableViewController和UILocalizedIndexedCollat​​ion

Question

我有一個表使用NSFetchedResultsController。這讓我與存在

- (NSInteger)numberOfSectionsInTableView:(UITableView *)tableView 
{ 
    return [[_fetchedResultsController sections] count]; 
} 

-(NSString *)tableView:(UITableView *)tableView titleForHeaderInSection:(NSInteger)section 
{ 
    return [[[_fetchedResultsController sections] objectAtIndex:section] name]; 
} 

- (NSInteger)tableView:(UITableView *)tableView numberOfRowsInSection:(NSInteger)section 
{ 
    NSInteger numberOfRows = 0; 
    NSArray *sections = _fetchedResultsController.sections; 
    if(sections.count > 0) 
    { 
     id <NSFetchedResultsSectionInfo> sectionInfo = [sections objectAtIndex:section]; 
     numberOfRows = [sectionInfo numberOfObjects]; 
    } 
    return numberOfRows; 
} 

- (NSArray *)sectionIndexTitlesForTableView:(UITableView *)tableView 
{ 
    return [_fetchedResultsController sectionIndexTitles]; 
} 

- (NSInteger)tableView:(UITableView *)tableView sectionForSectionIndexTitle:(NSString *)title atIndex:(NSInteger)index 
{ 
    return [_fetchedResultsController sectionForSectionIndexTitle:title atIndex:index]; 
} 

頭的索引，但我想用UILocalizedIndexCollat​​ion以獲得完整的字母表。

如何將這些方法連接到NSFetchedResultsController？

我想我需要從目前的整理

- (NSArray *)sectionIndexTitlesForTableView:(UITableView *)tableView 
{ 
    return [[UILocalizedIndexedCollation currentCollation] sectionIndexTitles]; 
} 

獲得指數冠軍，但我失去了對如何寫這個方法

- (NSInteger)tableView:(UITableView *)tableView sectionForSectionIndexTitle:(NSString *)title atIndex:(NSInteger)index 
{ 
    //How do I translate index here to be the index of the _fetchedResultsController? 
    return [_fetchedResultsController sectionForSectionIndexTitle:title atIndex:index]; 
}  

Answer 1

我認爲你必須遍歷獲取的成果例如：

- (NSInteger)tableView:(UITableView *)tableView sectionForSectionIndexTitle:(NSString *)title atIndex:(NSInteger)index 
{ 
    NSInteger section = 0; 
    for (id <NSFetchedResultsSectionInfo> sectionInfo in [_fetchedResultsController sections]) { 
     if ([sectionInfo.indexTitle compare:title] >= 0) 
      break; 
     section++; 
    } 
    return section; 
} 

對於沒有匹配部分的部分索引標題，您必須決定是否要跳轉到「較低」或「較高」部分。上面的方法跳轉到下一個更高的部分。

Answer 2

我從Martin那裏得到的解決方案。

- (NSInteger)tableView:(UITableView *)tableView sectionForSectionIndexTitle:(NSString *)title atIndex:(NSInteger)index 
{ 
    NSInteger localizedIndex = [[UILocalizedIndexedCollation currentCollation] sectionForSectionIndexTitleAtIndex:index]; 
    NSArray *localizedIndexTitles = [[UILocalizedIndexedCollation currentCollation] sectionIndexTitles]; 
    for(int currentLocalizedIndex = localizedIndex; currentLocalizedIndex > 0; currentLocalizedIndex--) { 
     for(int frcIndex = 0; frcIndex < [[_fetchedResultsController sections] count]; frcIndex++) { 
      id<NSFetchedResultsSectionInfo> sectionInfo = [[_fetchedResultsController sections] objectAtIndex:frcIndex]; 
      NSString *indexTitle = sectionInfo.indexTitle; 
      if([indexTitle isEqualToString:[localizedIndexTitles objectAtIndex:currentLocalizedIndex]]) { 
       return frcIndex; 
      } 
     } 
    } 
    return 0; 
}