如何在 Objective-C iphone sdk 中迭代嵌套字典

Question

您好，我有一个 json 字符串，已通过 JSON 框架转换为字典，我需要提取其内容。我怎样才能迭代到嵌套字典？我已经有了这段代码，可以让我查看字典：

NSDictionary *results = [responseString JSONValue]; 
NSMutableArray *catArray = [NSMutableArray array];

for (id key in results) {
    NSLog(@"key: %@, value: %@", key, [results objectForKey:key]);
    [catArray addObject:key];
    NSString *cat = key;
}

有人可以提供一个示例，说明如何在不使用键名的情况下到达 dic 的所有级别吗？

dic的结果结构是这样的： http://www.freeimagehosting.net/uploads/ e7c020d697.png

替代文本 http://www.freeimagehosting.net/uploads/e7c020d697 .png

谢谢

ldj

Answer 1

我尝试过递归，希望这能有所帮助。
PS 这将返回该键的第一次出现。

+(id)getValue:(NSDictionary*)aDictionary forKey:(NSString *)aKey {
   id finalValue = nil;
   for (id key in [aDictionary allKeys]) {
      id value = aDictionary[key];
      if (value != nil && [key isEqualToString:aKey]) {
         finalValue = value;
         break;
      }
      else if ([value isKindOfClass:[NSDictionary class]]) {
       finalValue = [[self class] getValue:value forKey:aKey];
      }
   }
   return finalValue;
}

Answer 2

我确信您现在已经解决了这个问题，但我发现编写一个递归实用程序类函数来递归记录字典的每个键很方便：

/**
 *  Recursively logs the keys and values of each object in the dictionary,
 *  along with the class of the value.
 *
 *  @param dict The dictionary to inspect
 */
+ (void)logDictionaryKeys:(NSDictionary*)dict {
    for (id key in dict) {
        if ([dict[key] isKindOfClass:[NSDictionary class]]) {
            [self logDictionaryKeys:dict[key]];
        }else {
            NSLog(@"Key: %@", key);
            NSLog(@"Value: %@ (%@)", dict[key], [dict[key] class]);
        }
    }

    return;
}

Answer 3

类别的标题：

#import <Foundation/Foundation.h>

@interface NSDictionary(Find)

-(id) findflat:(NSString*)keyToFind;

@end

正文：用法

#import "NSDictionary+Find.h"

@implementation NSDictionary(Find)

-(id) findflat:(NSString*)keyToFind{

    if([self objectForKey:keyToFind])
        return self[keyToFind];

    for(id key in self)
        if([[self objectForKey:key] isKindOfClass:[NSDictionary class]])
            return [[self objectForKey:key] findflat:keyToFind];


    return nil;
}

@end

/验证：

- (void)testExample {
// This is an example of a functional test case.
// Use XCTAssert and related functions to verify your tests produce the correct results.

NSDictionary *dic = @{};
XCTAssertNil([dic findflat:@"url"]);

dic = @{@"url":@"http://"};
XCTAssertEqual([dic findflat:@"url"],@"http://");

dic = @{@"other":@4};
XCTAssertNil([dic findflat:@"url"]);

dic = @{@"other":@4,@"url":@"http://"};
XCTAssertEqual([dic findflat:@"url"],@"http://");

dic = @{@"other":@4,@"sd":@"sdf"};
XCTAssertNil([dic findflat:@"url"]);

dic = @{@"other":@4,@"dic":@{@"url":@"http://"}};
XCTAssertEqual([dic findflat:@"url"],@"http://");

dic = @{@"other":@4,@"dic":@{@"sd":@2,@"url":@"http://"}};
XCTAssertEqual([dic findflat:@"url"],@"http://");

dic = @{@"other":@4,@"dic":@{}};
XCTAssertNil([dic findflat:@"url"]);

dic = @{@"other":@4,@"dic":@{@"dic":@{@"sd":@2,@"url":@"http://"}}};
XCTAssertEqual([dic findflat:@"url"],@"http://");

}

Answer 4

您可能必须尝试递归函数。

-(void)recurse:(NSDictionary *)dict counter: (int*) i parent:(NSString *) parent{
    for (NSString* key in [dict allKeys]) {
        id value = [dict objectForKey:key];
        NSLog(@"%@ -> %@", parent, key);
        if ([value isKindOfClass:[NSDictionary class]]) {
            i++;
            NSDictionary* newDict = (NSDictionary*)value;
            [self recurse:newDict counter:i parent:key];
            i--;
        } else {
            //Do smthg
        }
    }
}

这将列出所有键及其级别。

Answer 5

尝试

for ((id) key in [results allKeys]) {
    NSLog(@"key: %@, value: %@", key, [results objectForKey:key]);
    [catArray addObject:key];
    NSString *cat = key;
}

希望有帮助。

编辑：
我不确定我是否理解你的结构正确，但我假设你的字典中有一些值，它们又是字典。
如果你想通过位于另一个字典内部的字典进行交互，你可以这样做：

for ((id) key in [results allKeys]) {
    id value = [results objectForKey:key];
    if ([value isKindOfClass:[NSDictionary class]]) {
        NSDictionary* newDict = (NSDictionary*)value;
        for ((id) subKey in [newDict allKeys]) {
            ...
        }
    }
}

当然，如果你知道所需字典的特定键，你可以检查它而不是检查该值是否是字典。但也许在这两种情况下检查一下并不是一个坏主意......