NSPredicateを使用して検索 - フルネーム

Question

これはほぼ3日間続きましたが、私はまだバグを修正できません。シンプルな状況で、私は人の名前（名前と姓）を検索したいと思います。

例：

その後、私は（結果：何も）の後に空白でジョンを書く私は続ける。その後

：

1. まず、私はジョン（ジョン・ニューマン、安定ジョン、ジョンCarensで結果を）書きますジョンNと（結果：ジョンニューマン）。検索に空白を入れても、 の名前を表示する必要があります。

これはif/elseの最初の部分で、2+語の配列で動作します。おかげ

- (void)searchForText:(NSString*)searchText 
{ 
NSString *searchTextt = [searchText stringByTrimmingCharactersInSet:[NSCharacterSet whitespaceCharacterSet]]; 

     NSArray *array = [searchTextt componentsSeparatedByString:@" "]; 


     NSString *firstName = searchTextt; 
     NSString *lastName = searchTextt; 
     NSString *firstName2 = searchTextt; 
     NSString *lastName2 = searchTextt; 

     NSPredicate *predicate = nil; 

    if ([array count] > 1) { 
     firstName = array[0]; 
     lastName = array[1]; 
     firstName2 = array[1]; 
     lastName2 = array[0]; 

     predicate = [NSPredicate predicateWithFormat:@"(guestName CONTAINS[cd] %@ AND guestSurname CONTAINS[cd] %@)", firstName, lastName]; 
     NSPredicate *predicate2 = [NSPredicate predicateWithFormat:@"(guestSurname CONTAINS[cd] %@ AND guestName CONTAINS[cd] %@)", lastName2, firstName2]; 

     NSPredicate *compoundPredicate = [NSCompoundPredicate orPredicateWithSubpredicates:@[predicate, predicate2]]; 


     self.searchResults = [self.mainarray filteredArrayUsingPredicate:compoundPredicate]; 

    } else { 
     NSPredicate *predicate = [NSPredicate predicateWithFormat:@"guestSurname contains[cd] %@", searchText]; 
     NSPredicate *predicate2 = [NSPredicate predicateWithFormat:@"guestName contains[cd] %@", searchText]; 
     NSPredicate *predicate3 = [NSPredicate predicateWithFormat:@"guestCompany contains[cd] %@", searchText]; 


     NSPredicate *compoundPredicate = [NSCompoundPredicate orPredicateWithSubpredicates:@[predicate, predicate2, predicate3]]; 
     self.searchResults = [self.mainarray filteredArrayUsingPredicate:compoundPredicate]; 

    } 


} 

EDIT

---

：

- (void)updateSearchResultsForSearchController:(UISearchController *)searchController 
{ 
    NSString *searchString = searchController.searchBar.text; 
    [self searchForText:searchString]; 
    [_maintable reloadData]; 
} 

Answer 1

私は

（様配列{ guestName =ジョン; guestSurname =ニューマン;}、{ guestName = John; gue stSurname =安定です。 }、{ guestName = John; guestSurname = Carens; }）

私はのfuction

- (void)searchForText:(NSString*)searchText 
    { 
     searchTextt = [searchText stringByTrimmingCharactersInSet:[NSCharacterSet whitespaceCharacterSet]]; 
     NSArray *array = [searchTextt componentsSeparatedByString:@" "]; 
     NSString *firstName = searchTextt; 
     NSString *lastName = searchTextt; 
     NSPredicate *predicate = nil; 

     if ([array count] > 1) { 
      firstName = array[0]; 
      lastName = array[1]; 
      predicate = [NSPredicate predicateWithFormat:@"(guestName CONTAINS[cd] %@ AND guestSurname CONTAINS[cd] %@) OR (guestName CONTAINS[cd] %@ AND guestSurname CONTAINS[cd] %@)", firstName, lastName, lastName, firstName]; 

     } else { 
      predicate = [NSPredicate predicateWithFormat:@"guestName CONTAINS[cd] %@ OR guestSurnamen CONTAINS[cd] %@", firstName, lastName]; 

     } 
     NSArray *arrResult = [arr filteredArrayUsingPredicate:predicate];; 

    } 

- (void)updateSearchResultsForSearchController:(UISearchController 
*)searchController 
{ 
    NSString *searchString = searchController.searchBar.text; 
    [self searchForText:searchString]; 
    [_maintable reloadData]; 
} 

を書くには、これを試してみてください。テキストフィールドに「john」と入力すると結果が得られます。