Question

我有一个字典的NSMutableArray。我正在使用NSPredicate来过滤数组，以查找是否存在具有特定键的字典。

我已经提到了各种各样的例子，其中最接近的是：Using NSPredicate to filter an NSArray based on NSDictionary keys。但是，我不希望钥匙有价值。我的问题是我想先找到钥匙。我尝试了不同的语法，但没有帮助。

到目前为止我做了什么：

    NSString *key = @"open_house_updated_endhour";
    NSPredicate *predicateString = [NSPredicate predicateWithFormat:@"%K", key];
    //NSPredicate *predicateString = [NSPredicate predicateWithFormat:@"%K contains [cd]", key]; Doesn't work.
    //NSPredicate *predicateString = [NSPredicate predicateWithFormat:@"%K == %@", key]; Won't work because it expects a value here.
    NSLog(@"predicate %@",predicateString);
    NSArray *filtered = [updatedDateAndTime filteredArrayUsingPredicate:predicate]; // updatedDateAndTime is the NSMutableArray

Answer 1

字典提供nil作为缺席密钥的值。因此，只需将密钥与nil进行比较。

NSPredicate *predicateString = [NSPredicate predicateWithFormat:@"%K==NULL", key]; // Dictionaries not having the key
NSPredicate *predicateString = [NSPredicate predicateWithFormat:@"%K!=NULL", key]; // Dictionaries having the key

Answer 2

试一试：

 NSArray *Myarray = [NSArray arrayWithObject:[NSMutableDictionary dictionaryWithObject:@"my hello string" forKey:@"name"]];   
NSArray *filteredarray = [Myarray filteredArrayUsingPredicate:[NSPredicate predicateWithFormat:@"(name == %@)", @"my hello string"]];
NSLog("%@",filteredarray);

另一个例子：

NSString *mycategory = @"iamsomeone";
NSArray *myitems = @[@{ @"types" : @[@"novel", @"iamsomeone", @"dog"] },
                   @{ @"types" : @[@"cow", @"iamsomeone-iam", @"dog"] },
                   @{ @"types" : @[@"cow", @"bow", @"cat"] }];
NSPredicate *mypredicate = [NSPredicate predicateWithBlock:^BOOL(id evaluatedObject, NSDictionary *bindings) {
    NSArray *categories = [evaluatedObject objectForKey:@"types"];
    return [categories containsObject:mycategory];
}];

NSArray *outpuArray = [myitems filteredArrayUsingPredicate:mypredicate];
NSLog(@"hello output:%@",outpuArray);

使用NSPredicate过滤NSDictionaries的NSMutableArray以查找＆＃34;密钥＆＃34;存在

2 个答案: