使用NSPredicate基于多个键进行过滤(键的NOT值)

Question

我有以下包含NSDictionary的NSArray:

NSArray *data = [[NSArray alloc] initWithObjects:
                 [NSDictionary dictionaryWithObjectsAndKeys:[NSNumber numberWithInt:1], @"bill", [NSNumber numberWithInt:2], @"joe", nil],
                 [NSDictionary dictionaryWithObjectsAndKeys:[NSNumber numberWithInt:3], @"bill", [NSNumber numberWithInt:4], @"joe", [NSNumber numberWithInt:5], @"jenny", nil],
                 [NSDictionary dictionaryWithObjectsAndKeys:[NSNumber numberWithInt:6], @"joe", [NSNumber numberWithInt:1], @"jenny", nil],
                 nil];

我想创建一个过滤的NSArray,它只包含NSDictionary使用NSPredicate匹配多个"键"的对象.

例如:

• 过滤数组只包含具有键"bill"和"joe"的NSDictionary对象[所需结果:新的NSArray将包含前两个NSDictionary对象]
• 过滤数组只包含具有键"joe"和"jenny"的NSDictionary对象[所需结果:新的NSArray将包含最后两个NSDictionary对象]

任何人都可以解释NSPredicate的格式来实现这一目标吗？

编辑:我可以使用以下方法获得与所需NSPredicate类似的结果:

NSMutableArray *filteredSet = [[NSMutableArray alloc] initWithCapacity:[data count]];
NSString *keySearch1 = [NSString stringWithString:@"bill"];
NSString *keySearch2 = [NSString stringWithString:@"joe"];

for (NSDictionary *currentDict in data){
    // objectForKey will return nil if a key doesn't exists.
    if ([currentDict objectForKey:keySearch1] && [currentDict objectForKey:keySearch2]){
        [filteredSet addObject:currentDict];
    }
}

NSLog(@"filteredSet: %@", filteredSet);

我想象如果存在NSPredicate会更优雅吗？

Answer 1

他们唯一知道的方法是结合两个条件,如"'value1'IN列表和'value2'IN列表"

self.@ allKeys应该返回字典中的所有键(self是数组中的每个字典).如果你不用前缀@写它,那么字典只会寻找一个"allKeys"而不是方法" - (NSArray*)allKeys"

代码:

NSArray* billAndJoe = [data filteredArrayUsingPredicate:[NSPredicate predicateWithFormat:@"%@ IN self.@allKeys AND %@ IN self.@allKeys" , @"bill",@"joe" ]];


NSArray* joeAndJenny = [data filteredArrayUsingPredicate:[NSPredicate predicateWithFormat:@"%@ IN self.@allKeys AND %@ IN self.@allKeys" , @"joe",@"jenny" ]]

Answer 2

因为nil如果你要求一个不存在的键的值,字典就会返回,所以指定该值应该是非零是足够的.以下格式应涵盖您的第一种情况:

[NSPredicate predicateWithFormat: @"%K != nil AND %K != nil", @"bill", @"joe"]

第二种情况,"joe"和"jenny"遵循类似的模式,当然.