如何使用多个条件筛选托管对象实体

Question

我可能会以错误的方式思考这个问题。

我有一个处理实体的搜索/过滤功能。但我希望能够过滤几个标准..

我有这个：

for (WordEntity *word in [self.fetchedResultsController fetchedObjects])
{
    if ([scope isEqualToString:@"All"] || [word.greekText isEqualToString:searchText ])
    {
        NSComparisonResult result = [word.greekText compare:searchText
                                               options:(NSCaseInsensitiveSearch|NSDiacriticInsensitiveSearch)
                                                 range:NSMakeRange(0, [searchText length])];
        if (result == NSOrderedSame)
        {

            [self.searchResults addObject:word];
        }
    }

但我希望过滤器捕获英文和希腊文字。我怎么做到最好？感谢

已更新

道歉，但我仍然无法让它发挥作用..我已经打开了它..

正如所建议的那样，我按如下方式更改了代码：

 [self.searchResults removeAllObjects];
  NSLog(@"predicating");

NSPredicate *englishTextPredicate = [NSPredicate predicateWithFormat:@"englishText CONTAINS[cd] %@", searchText];
NSLog(@"eng pred: %@", englishTextPredicate);

NSLog(@"search text: %@", searchText);


 [self.fetchedResultsController.fetchRequest setPredicate:englishTextPredicate];
 NSArray *fetchedData = [self.fetchedResultsController fetchedObjects];

 self.searchResults = [fetchedData copy];
 NSLog(@"search count: %i", self.searchResults.count);

但这会产生以下输出。

2014-08-03 07:31:40.846 GuessGreek[16535:425669] Card Count is: 651
2014-08-03 07:31:54.495 GuessGreek[16535:425669] -[WordViewController searchDisplayController:shouldReloadTableForSearchString:]
2014-08-03 07:31:54.495 GuessGreek[16535:425669] -[WordViewController filterContentForSearchText:scope:]
2014-08-03 07:31:54.496 GuessGreek[16535:425669] predicating
2014-08-03 07:31:54.496 GuessGreek[16535:425669] eng pred: englishText.text CONTAINS[cd] "o"
2014-08-03 07:31:54.496 GuessGreek[16535:425669] search text: o
2014-08-03 07:31:54.497 GuessGreek[16535:425669] search count: 651

这是单词的截图..显然，前两个单词没有＆＃34; o＆＃34;

我做错了什么？

Answer 1

使用谓词会更简单。您可以使用复合谓词来搜索两件事。

NSPredicate * pred1 = [NSPredicate predicateWithFormat:@"greekText == %@", searchText];
NSPredicate * pred2 = [NSPredicate predicateWithFormat:@"englishText == %@", searchText];
NSPredicate * compPred = [NSCompoundPredicate andPredicateWithSubpredicates:[NSArray arrayWithObjects:pred1,pred2,nil]];
[YourFetchRequestName setPredicate:compPred];

然后执行你的获取请求只能提供你想要的东西。

self.searchResults = [yourContext executeFetchRequest:YourFetchRequestName error:&error];

希望有所帮助！

Answer 2

我发现一些效果很好的我认为：

-(void)filterContentForSearchText:(NSString*)searchText scope:(NSString*)scope {
    [self.searchResults removeAllObjects];
   NSLog(@"%s", __FUNCTION__);
    for (WordEntity *word in [self.fetchedResultsController fetchedObjects])
    {
    NSRange greekResult  = [word.greekText rangeOfString:searchText options:NSCaseInsensitiveSearch|NSDiacriticInsensitiveSearch];
    NSRange englishResult = [word.englishText rangeOfString:searchText options:NSCaseInsensitiveSearch|NSDiacriticInsensitiveSearch];

        if ( greekResult.location != NSNotFound || englishResult.location != NSNotFound ) {

            [self.searchResults addObject:word];
        }