Question

我有三个领域对象。部门，部门和用户。 section是某个部门的一个子部门。但是每个部门和每个部门都会有用户。

@interface Department : RLMObject

@property NSString *name;
@property BOOL isCollapsed;

@property (nonatomic, strong) RLMArray<Section> *sections;
@property (nonatomic, strong) RLMArray<User> *users;

@end

@interface Section : RLMObject

@property NSString *name;
@property BOOL isCollapsed;
@property RLMArray<User> *users;

@end 

@interface User : RLMObject

@property NSString *department;
@property NSString *email;
@property NSString *firstname;
@property NSString *lastname;
@property NSString *fullname;

@end

我想要做的是，我想搜索名字和姓氏包含的所有用户＆＃34; a＆＃34;并与部门和部门联系。

现在我正在做的是

searchText = [searchText lowercaseString];

RLMResults *sections = [Section objectsWhere:
                        [NSString stringWithFormat:@"SUBQUERY(users, $user, $user.fullname contains '%@' OR $user.nickname contains '%@').@count > 0",searchText,searchText]];

NSMutableArray *sectionNames = [NSMutableArray array];

for (Section *section in sections)
{
    [sectionNames addObject:section.name];
}

NSPredicate *predicate = [NSPredicate predicateWithFormat:@"SUBQUERY(sections, $section, $section.name IN %@).@count > 0", sectionNames];

RLMResults *filteredDepartments = [[Department objectsWithPredicate:predicate] sortedResultsUsingProperty:@"name" ascending:YES];

NSPredicate *pred = [NSPredicate predicateWithFormat:@"SUBQUERY(users, $user, $user.fullname contains %@ OR $user.nickname contains %@).@count > 0",searchText,searchText];
RLMResults *departments = [filteredDepartments objectsWithPredicate:pred];

目前问题是部分数组中有两个部分，但在部署结果中，它返回null。请帮忙。谢谢。

注意：如果一个用户已经属于某个部分，那么该用户将不会包含在部门中。

Answer 1

也许您不需要使用子查询。更简单地说，您可以在查询中使用ANY，如下所示：

[Department objectsWhere:
    @"ANY users.firstname CONTAINS %@ OR ANY users.lastname CONTAINS %@ OR ANY sections.users.firstname CONTAINS %@ OR ANY sections.users.lastname CONTAINS %@", searchText, searchText, searchText, searchText];

但我认为使用反向关系更容易。在Realm中，反向关系定义为RLMLinkingObjects。您可以按如下方式向User类添加反向关系：

@interface User : RLMObject

...

@property (readonly) RLMLinkingObjects *departments;
@property (readonly) RLMLinkingObjects *sections;

@end

@implementation User

+ (NSDictionary *)linkingObjectsProperties {
    return @{@"departments": [RLMPropertyDescriptor descriptorWithClass:Department.class propertyName:@"users"],
             @"sections": [RLMPropertyDescriptor descriptorWithClass:Section.class propertyName:@"users"]};
}

@end

然后，您可以从User的属性获取用户所属的部门和部分，如下所示：

RLMResults *users = [User objectsWhere:@"firstname CONTAINS %@ OR lastname CONTAINS %@" , searchText, searchText];
for (User *user in users) {
    NSLog(@"%@", user.departments);
    NSLog(@"%@", user.sections);
}

现实中的子查询和IN语句

1 个答案: