使用NSPredicate OR语句进行搜索的最有效方法 [英] most effective way of searching using NSPredicate OR statement

问题描述

我试图在上一篇文章中建立一个谓词，但最终这样做是因为3个小时后，我无法正常工作。

I tried to build a predicate in a previous post, but ended up doing it this way because after 3 hours, I couldn't get it to work.

我觉得必须有一种更有效的方法，还需要谓词对大小写不敏感。

I feel like there has to be a more effective way, also I need the predicates to be case insensitive.

10,000辆车
我有轮胎，车轮，作为三个汽车零件的座位
我想找到所有有轮胎的汽车。
然后，我想找到所有有轮毂的汽车。
然后我想找到所有有座位的汽车。
（我知道很多重复项，但这就​​是我所需要的）。

10,000 cars I have a tire, wheel, seat as three car parts I want to find all cars that have a tire. THEN I want to find all cars that have a wheel. THEN I want to find all cars that have a seat. (I know many will be duplicates, but that's what I need)

请告诉我是否有更有效的方法。
也请让我知道如何使谓词不区分大小写。

Please let me know if there is a more effective way. Also please let me know how to make the predicates case insensitive.

谢谢！

-(NSArray*) loadCarsFromCoreData:(NSMutableArray*)inputCarParts{

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

    //To find the cars we are using the car parts
    NSEntityDescription *entity = [NSEntityDescription entityForName:@"CarParts" inManagedObjectContext:[self managedObjectContext]];
    [fetchRequest setEntity:entity];
    NSError *error;
    NSMutableArray *predicates =[self parseCarPartsIntoAPredicateArray:inputCarParts];
    for (int i=0; i<[predicates count]; i++) {
        [fetchRequest setPredicate:[predicates objectAtIndex:i]];

        NSArray *records = [[self managedObjectContext] executeFetchRequest:fetchRequest error:&error];
        NSLog(@"results = %i",[records count]);
        [totalResults addObjectsFromArray:records];
    }
    NSLog(@"results = %i",[totalResults count]);
    return [NSArray arrayWithArray:totalResults];
}

-(NSMutableArray*)parseCarPartsIntoAPredicateArray:(NSMutableArray*)inputCarParts{
    NSMutableArray *returnArray=[[NSMutableArray alloc]init];
    for (int i=0; i<[inputCarParts count]; i++) {
        NSPredicate *predicate=[NSPredicate predicateWithFormat:@"partName == %@",[inputCarParts objectAtIndex:i]];
        [returnArray addObject:predicate];
    }

    return returnArray;
}

推荐答案

听起来像你真正要寻找的是如何根据可能的匹配项数组构造谓词。在这种情况下，您可以执行以下操作：

It sounds like what you're really looking for is how to construct a predicate based on an array of possible matches. In that case you'd do something like this:

NSMutableArray *partPredicates = [NSMutableArray arrayWithCapacity:[inputCarParts count]];
for (NSString *partName in inputCarParts) {
    NSPredicate *currentPartPredicate = [NSPredicate predicateWithFormat:@"partName =[c] %@", partName];
    [partPredicates addObject:currentPartPredicate];
}
NSPredicate *fullPredicate = [NSCompoundPredicate orPredicateWithSubpredicates:partPredicates];

然后得出的谓词类似于 partName == [c]轮胎或部件名称== [c]车轮或部件名称== [c]座椅 ，每个数组条目具有一个组件。一次提取即可一次评估它们。

The resulting predicate would then be something like partName ==[c] "tire" OR partName ==[c] "wheel" OR partName ==[c] "seat", with one component per array entry. Do one fetch to evaluate them all in one shot.

如果在提取中使用它，将不会得到重复的结果。您对这个问题发表了评论，表明您不想重复。如果是这种情况，那么我不确定我知道很多重复的东西，但这就是我需要的。听起来好像您想要骗子，但您却说自己没有。

This will not get duplicate results if you use it in the fetch. You left a comment on the question that indicates you don't want duplicates. If that's the case though, then I'm not sure what I know many will be duplicates, but that's what I need means. It sounds like you wanted the dupes, but you said you didn't.

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