Question

我想知道是否有一种方法可以简化NSPredicate，它可以为多个比较目标提供单个查询字符串。我正在为同一个查询字符串搜索核心数据实体的多个属性。我当前的查询看起来像这样......

[NSPredicate predicateWithFormat:@"(attributeA contains[cd] %@) OR (attributeB contains[cd] %@) OR (attributeC contains[cd] %@)", searchString, searchString, searchString];

请注意，这样做效果很好，但确实看起来有点不雅观。特别是searchString, searchString, searchString部分。关于如何简化这一点的任何提示都会很棒！

谢谢！

Answer 1

您可以将NSCompoundPredicate用于您的OR＆amp;和这样的操作。

Obj-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

NSPredicate *predicate1 = [NSPredicate predicateWithFormat:@"X == 1"];
NSPredicate *predicate2 = [NSPredicate predicateWithFormat:@"X == 2"];
NSPredicate *predicate = [NSCompoundPredicate andPredicateWithSubpredicates:@[predicate1, predicate2]];

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

let predicate1:NSPredicate = NSPredicate(format: "X == 1")
let predicate2:NSPredicate = NSPredicate(format: "Y == 2")
let predicate:NSPredicate  = NSCompoundPredicate(andPredicateWithSubpredicates: [predicate1,predicate2] )

Swift 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

    let predicate1 = NSPredicate(format: "X == 1")
    let predicate2 = NSPredicate(format: "Y == 2")
    let predicateCompound = NSCompoundPredicate.init(type: .and, subpredicates: [predicate1,predicate2])

Answer 2

你可以这样做：

NSPredicate *p = [NSPredicate predicateWithFormat:@"attributeA contains[cd] $A OR attributeB contains[cd] $A or attributeC contains[cd] $A"];
NSDictionary *sub = [NSDictionary dictionaryWithObject:searchString forKey:@"A"];
p = [p predicateWithSubstitutionVariables:sub];

或者你可以做一些更奇怪的事情，比如：

- (NSPredicate *)buildOrPredicate:(NSDictionary *)stuff {
  NSMutableArray *subs = [NSMutableArray array];
  for (NSString *key in stuff) {
    NSString *value = [stuff objectForKey:stuff];
    NSPredicate *sub = [NSPredicate predicateWithFormat:@"%K contains[cd] %@", key, value];
    [subs addObject:sub];
  }
  return [NSCompoundPredicate orPredicateWithSubpredicates:subs];
}

然后用：

调用它

NSDictionary *stuff = [NSDictionary dictionaryWithObjectsAndKeys:
                       searchString, @"attributeA",
                       searchString, @"attributeB",
                       searchString, @"attributeC",
                       nil];
NSPredicate *p = [self buildOrPredicate:stuff];

我能想到的唯一可能有用的是尝试在谓词格式中使用位置指定。但是，我不知道解析器是否以与+stringWithFormat:相同的方式识别它们：

NSPredicate *p = [NSPredicate predicateWithFormat:@"attributeA contains[cd] %1$@ OR attributeB contains[cd] %1$@ or attributeC contains[cd] %1$@", searchString];

NSPredicate与一个查询字符串进行多次比较

2 个答案: