使用NSPredicate的NSFetchRequest

Question

我正在尝试以这种方式通过核心数据数据库创建所有过滤搜索结果的NSArray：

NSFetchRequest *fetchRequest = [[NSFetchRequest alloc] init];

NSEntityDescription *entity = [NSEntityDescription
                               entityForName:@"AllFiles" inManagedObjectContext:context];
[fetchRequest setEntity:entity];

NSPredicate *p1 = [NSPredicate predicateWithFormat:@"userID == %@", userID];
NSPredicate *p2 = [NSPredicate predicateWithFormat:@"folderNumber == %d", folderNumber];
NSPredicate *p3 = [NSPredicate predicateWithFormat:@"fileType == %@", type];

NSPredicate *fetchPredicate = [NSCompoundPredicate orPredicateWithSubpredicates:@[p1, p2, p3]];
[fetchRequest setPredicate:fetchPredicate];

NSError *error;
NSArray* returnArray = [context executeFetchRequest:fetchRequest error:&error];

这些文件是这样创建的：

AllFiles *newFile = [NSEntityDescription
                                        insertNewObjectForEntityForName:@"AllFiles"
                                        inManagedObjectContext:context];
newFile.userID = userID;
newFile.folderNumber = [NSString stringWithFormat:@"%d", folderNumber];
....

正如您所看到的，我正在尝试过滤给定的一些参数的结果：

@"userID == %@", userID

但是，返回所有结果，而不进行过滤...你知道为什么吗？

Answer 1

您正在使用OR谓词，这3个谓词中至少有一个与所有谓词匹配。我怀疑[NSPredicate predicateWithFormat:@"fileType == %@", type]在收集你所有的文件

试着像这样合成一个和...

NSPredicate *idPredicate = [NSPredicate predicateWithFormat:@"userID == %@", userID];
NSPredicate *folderNumPredicate = [NSPredicate predicateWithFormat:@"folderNumber == %d", folderNumber];
NSPredicate *typePredicate = [NSPredicate predicateWithFormat:@"fileType == %@", type];

NSPredicate *fetchPredicate = [NSCompoundPredicate andPredicateWithSubpredicates:@[idPredicate, folderNumPredicate, typePredicate];

[fetchRequest setPredicate:fetchPredicate];

请记住，andPredicateWithSubpredicates和orPredicateWithSubpredicates会返回一个NSPredicate，因此如果需要，可以将它们以复杂的方式组合在一起。