ios 在 NSPredicate 中结合“AND”和“OR”条件
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Combining 'AND' and 'OR' Condition in NSPredicate
提问by dubbeat
Back again needing more help with constructing my NSPredicates :(
再次需要更多帮助来构建我的 NSPredicates :(
Category
{
name:string
subs<-->>SubCategory
}
SubCategory
{
name:string
numbervalue:NSNumber
}
I would like to know how to create a predicate with ANDand OR.
我想知道如何使用AND和OR创建谓词。
For example, I would like to return every Category with name == "category_name" that also has a subcategory with a numbervalue of "valueA" OR"valueB".
例如，我想返回名称 == "category_name" 的每个类别，该类别还有一个数字值为 "valueA"或"valueB"的子类别。
I've tried every possible combination of predicate constructors I could come up with but I just cant get it to work.
我已经尝试了我能想到的所有可能的谓词构造函数组合，但我就是无法让它工作。
This is my best attempt so far.
这是我迄今为止最好的尝试。
NSPredicate *predicate=[NSPredicate predicateWithFormat@"(name== %@) AND (ANY subs.numbervalue==%@ OR ANY subs.numbervalue==%@)", @"aname", value1, value2];
Edit 1
编辑 1
Second attempt. The filtering on the 'numbervalue' is still not working.
第二次尝试。对“数值”的过滤仍然无效。
NSNumber valueA=[NSNumber numberWithInt:20];
NSNumber valueA=[NSNumber numberWithInt:90];
NSArray *arr=[NSArray arrayWithObject:valueA,valueB,nil];
predicate=[NSPredicate predicateWithFormat:@"(name==%@)AND(ANY subs.numbervalue IN %@)",@"aname",arr];
Edit 2
编辑 2
I've tried filtering just by numberValue and have left the name out alltogether.
我试过仅按 numberValue 进行过滤，并将名称完全排除在外。
1) using this results in the entire set being returned even if only 1 item in the set has the value.
1) 使用此结果会返回整个集合，即使集合中只有 1 个项目具有该值。
predicate=[NSPredicate predicateWithFormat:@"(ANY subs.numbervalue IN %@)",arr];
2) Using this results in the same problem, every item in the set being returned even if only 1 of them matches.
2) 使用它会导致同样的问题，即使其中只有 1 个匹配，也会返回集合中的每个项目。
predicate=[NSPredicate predicateWithFormat:@"((SUBQUERY(subs, $x, $x.numbervalue == %@ or $x.numbervalue == %@).@count > 0)", value1, value2];
Also using the simplest version results in the same problem.... I didnt notice this earlier.
同样使用最简单的版本会导致同样的问题......我之前没有注意到这一点。
NSPredicate *predicate=[NSPredicate predicateWithFormat@"(ANY subs.numbervalue==%@ ,valueA];
So I think I have my problem narrowed down.
所以我想我的问题缩小了。
Category is an Entity. SubCategory is an Entity.
类别是一个实体。子类别是一个实体。
Category has a one to many relationship with SubCategory and is represented as an NSSet of SubCategory.
Category 与 SubCategory 是一对多的关系，表示为 SubCategory 的 NSSet。
Each SubCategory has an attribute named numbervalue.
每个 SubCategory 都有一个名为 numbervalue 的属性。
Edit 3
编辑 3
To test I have 2 Categorys. Both Categories have 10 subs, each with a numbervalue of 1 -10;
为了测试，我有 2 个类别。两个类别都有 10 个子项，每个子项的数值为 1 -10；
Using this code both categories are returned along with all 10 subs for each.
使用此代码返回两个类别以及每个类别的所有 10 个子项。
Expected result is both categories returned , each with just 2 subs.
预期结果是返回两个类别，每个类别只有 2 个子项。
I've traced out all my objects and the data is correct. The problem is I cant return a category that has filtered subs.
我已经找出了我所有的对象并且数据是正确的。问题是我无法返回已过滤子项的类别。
NSFetchRequest *fetchRequest = [[NSFetchRequest alloc] init];
NSManagedObjectContext *context=[[DataSource sharedDataSource]managedObjectContext];
NSEntityDescription *entity = [NSEntityDescription entityForName:@"Category" inManagedObjectContext:context];
[fetchRequest setEntity:entity];
[fetchRequest setFetchBatchSize:20];
NSSortDescriptor *sortDescriptor = [[NSSortDescriptor alloc] initWithKey:@"name" ascending:YES];
NSArray *sortDescriptors = [NSArray arrayWithObjects:sortDescriptor, nil];
NSPredicate *predicate;
NSNumber *a=[NSNumber numberWithFloat:1];
NSNumber *b=[NSNumber numberWithFloat:2];
predicate=[NSPredicate predicateWithFormat:@"(SUBQUERY(subs,$x,$x.numbervalue==%@ or $x.numbervalue==%@).@count> 0)",a,b];
[fetchRequest setPredicate:predicate];
[fetchRequest setSortDescriptors:sortDescriptors];
NSError *error = nil;
[NSFetchedResultsContoller deleteCacheWithName:nil];
NSFetchedResultsController *aFetchedResultsController = [[NSFetchedResultsController alloc] initWithFetchRequest:fetchRequest managedObjectContext:context sectionNameKeyPath:@"name" cacheName:@"Master"];
aFetchedResultsController.delegate = self;
self.fetchedResultsController = aFetchedResultsController;
if (![self.fetchedResultsController performFetch:&error]) {
NSLog(@"Unresolved error %@, %@", error, [error userInfo]);
abort();
}
采纳答案by Lorenzo B
I'll do the following using a SUBQUERY.
我将使用SUBQUERY.
[NSPredicate predicateWithFormat@"(name == %@) AND (SUBQUERY(subs, $x, $x.numbervalue == %@ or $x.numbervalue == %@).@count > 0)", @"aname", value1, value2];
Try it and let me know.
试试吧，让我知道。
Another solution could be to grab the whole collection subsand then filter the retrieved set with a predicate to check if elements match for value1or value2.
另一种解决方案可能是获取整个集合subs，然后使用谓词过滤检索到的集合以检查元素是否与value1或匹配value2。
Hope it helps.
希望能帮助到你。
Edit
编辑
If you follow Eimantassuggestion make to sure to create the right array:
如果您遵循Eimantas 的建议，请确保创建正确的数组：
NSNumber valueA = [NSNumber numberWithInt:20];
NSNumber valueB = [NSNumber numberWithInt:90];
NSArray *arr = [NSArray arrayWithObjects:valueA, valueB, nil];
回答by Onur Var
Addition to @Stuart's answer, you can use NSCompoundPredicate for your OR & AND operations like this.
除了@Stuart 的回答之外，您还可以将 NSCompoundPredicate 用于这样的 OR & AND 操作。
Obj-C - OR
对象-C - 或
// OR Condition //
NSPredicate *predicate1 = [NSPredicate predicateWithFormat:@"X == 1"];
NSPredicate *predicate2 = [NSPredicate predicateWithFormat:@"X == 2"];
NSPredicate *predicate = [NSCompoundPredicate orPredicateWithSubpredicates:@[predicate1, predicate2]];
Obj-C - AND
Obj-C - 与
NSPredicate *predicate1 = [NSPredicate predicateWithFormat:@"X == 1"];
NSPredicate *predicate2 = [NSPredicate predicateWithFormat:@"X == 2"];
NSPredicate *predicate = [NSCompoundPredicate andPredicateWithSubpredicates:@[predicate1, predicate2]];
Swift - OR
斯威夫特 - 或
let predicate1:NSPredicate = NSPredicate(format: "X == 1")
let predicate2:NSPredicate = NSPredicate(format: "Y == 2")
let predicate:NSPredicate = NSCompoundPredicate(orPredicateWithSubpredicates: [predicate1,predicate2] )
Swift - AND
斯威夫特 - 和
let predicate1:NSPredicate = NSPredicate(format: "X == 1")
let predicate2:NSPredicate = NSPredicate(format: "Y == 2")
let predicate:NSPredicate = NSCompoundPredicate(andPredicateWithSubpredicates: [predicate1,predicate2] )
Swift 3 - OR
斯威夫特 3 - 或
let predicate1 = NSPredicate(format: "X == 1")
let predicate2 = NSPredicate(format: "Y == 2")
let predicateCompound = NSCompoundPredicate.init(type: .or, subpredicates: [predicate1,predicate2])
Swift 3 - AND
斯威夫特 3 - 和
let predicate1 = NSPredicate(format: "X == 1")
let predicate2 = NSPredicate(format: "Y == 2")
let predicateCompound = NSCompoundPredicate.init(type: .and, subpredicates: [predicate1,predicate2])
回答by Stuart P.
Another option is to use an NSCompoundPredicate (note andPredicateWithSubpredicates does an AND but there are a few options to use, this should be much more efficient than doing 2 searches)
另一种选择是使用 NSCompoundPredicate （注意 andPredicateWithSubpredicates 执行 AND 但有几个选项可以使用，这应该比执行 2 次搜索更有效）
回答by zeeshan
For Swift, this is how you can do it:
对于 Swift，您可以这样做：
SWIFT 4.x - AND
SWIFT 4.x - 和
let p0 = NSPredicate(format: "X == 1")
let p1 = NSPredicate(format: "Y == 2")
fetchRequest.predicate = NSCompoundPredicate(andPredicateWithSubpredicates: [p0, p1])
SWIFT 4.x - OR
SWIFT 4.x - 或
let p0 = NSPredicate(format: "X == 1")
let p1 = NSPredicate(format: "Y == 2")
fetchRequest.predicate = NSCompoundPredicate(orPredicateWithSubpredicates: [p0, p1])
SWIFT 4.x - NOT
SWIFT 4.x - 不是
let p0 = NSPredicate(format: "X == 1")
let p1 = NSPredicate(format: "Y == 2")
fetchRequest.predicate = NSCompoundPredicate(notPredicateWithSubpredicates: [p0, p1])
Example
例子
This is an example from one of my codes, where I look for uid and sync:
这是我的一个代码中的一个示例，我在其中查找 uid 和 sync：
let p0 = NSPredicate(format: self.uid + " = '" + uid + "'")
let p1 = NSPredicate(format: self.sync + " = %@", NSNumber(value: true as Bool))
fetchRequest.predicate = NSCompoundPredicate(andPredicateWithSubpredicates: [p0, p1])
