Question

我有一个模特：

地方＆LT; ---＆GT;＆GT;类别

执行以下查询时：

NSFetchRequest *req = [NSFetchRequest fetchRequestWithEntityName:@"Category"];
if(![searchText isEqualToString:@""] && searchText != nil){
    req.predicate = [NSPredicate predicateWithFormat:@"(places.placeTitle contains %@) OR (places.details contains %@) OR (name contains %@)", searchText, searchText, searchText];
    NSLog(@"ModelHelper.getResults: Search text given, creating predicate: %@", req.predicate);
}
req.sortDescriptors = [NSArray arrayWithObject:[NSSortDescriptor sortDescriptorWithKey:@"name"
                                                                             ascending:YES
                                                                              selector:@selector(localizedCaseInsensitiveCompare:)]];
NSFetchedResultsController *controller = [[NSFetchedResultsController alloc]initWithFetchRequest:req
                                                                            managedObjectContext:context
                                                                              sectionNameKeyPath:nil
                                                                                       cacheName:nil];
NSError *error = nil;
[controller performFetch:&error];

我收到不应该存在的匹配项，例如：

searchText = "1234"

第一个结果是：

category[0].name = "Bars"
category[0].places[0].placeTitle = "Aaa"
category[0].places[0].details = "Aaa"
category[0].places[1].placeTitle = "123456"
category[0].places[1].details = "123456"
category[0].places[2].placeTitle = "test"
category[0].places[2].details = "test"

我的谓词是否有问题？感谢。

Answer 1

来自documentation：

重要说明：调用此方法后，不得修改fetchRequest。例如，您不得更改其谓词或排序顺序。

移动此段：

if(![searchText isEqualToString:@""] && searchText != nil){
    req.predicate = [NSPredicate predicateWithFormat:@"(places.placeTitle contains[c] %@) OR (places.details contains[c] %@) OR (name contains[c] %@)", searchText, searchText, searchText];
    NSLog(@"ModelHelper.getResults: Search text given, creating predicate: %@", req.predicate);
}

在初始化FRC之前

iOS Simple Predicate'包含'不起作用

1 个答案: