ios - 使用 NSPredicate 搜索 - 全名

Question

已经将近 3 天了，我仍然无法修复该错误。简单的情况，我想搜索人名(姓名和姓氏)。

示例:

1. 首先我写给 John(结果:John Newman、John Stable、John Carens)
2. 然后我用空格写 John(结果:无)
3. 之后我继续与 John N(结果:John Newman)合作。我需要保持即使我在搜索中输入空格，也会显示名称。

这只是 if/else 的第一部分，我在其中处理 2 个以上单词的数组。谢谢

---

- (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];

    }


}

编辑:

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

Answer 1

我有一个类似的数组

( { guestName = John; guestSurname = Newman; }, { guestName = John; guestSurname = Stable; }, { guestName = John; guestSurname = Carens; } )

我写函数

- (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];
}

试试这个。这对我有用，当我在文本字段中输入“john”时，我得到了结果。