Question

我需要合并两个字典而不重复

第一个喜欢：

{
  "Posts": [
    {
      "id": "1",
      "Title": "First post",
      "Category": "News"
    },
    {
      "id": "2",
      "Title": "Second post",
      "Category": "News"
    },
    {
      "id": "3",
      "Title": "Third post",
      "Category": "News"
    }
  ]
}

第二个像：

{
  "Posts": [
    {
      "id": "3",
      "Title": "Third post",
      "Category": "News"
    },
        {
      "id": "4",
      "Title": "Fourth post",
      "Category": "News"
    },
        {
      "id": "5",
      "Title": "Fifth post",
      "Category": "News"
    }
  ]
}

如何在不重复第三个条目的情况下合并？

任何人都可以提供帮助

Answer 1

  NSDictionary *dic1 = @{
          @"Posts" : @[
                  @{
                          @"id" : @"1", @"Title" : @"First post", @"Category" : @"News"
                  }, @{
                          @"id" : @"2", @"Title" : @"Second post", @"Category" : @"News"
                  }, @{
                          @"id" : @"3", @"Title" : @"Third post", @"Category" : @"News"
                  }
          ]
  };

  NSDictionary *dic2 = @{
          @"Posts" : @[
                  @{
                          @"id" : @"3", @"Title" : @"Third post", @"Category" : @"News"
                  }, @{
                          @"id" : @"4", @"Title" : @"Fourth post", @"Category" : @"News"
                  }, @{
                          @"id" : @"5", @"Title" : @"Fifth post", @"Category" : @"News"
                  }
          ]
  };

  NSMutableArray *keys = [NSMutableArray new];

  [dic1 enumerateKeysAndObjectsUsingBlock:^(id key, id obj, BOOL *stop) {

      [keys addObject:dic1[ key ]];
  }];

  [dic2 enumerateKeysAndObjectsUsingBlock:^(id key, id obj, BOOL *stop) {

      [keys addObject:dic2[ key ]];
  }];

  NSMutableArray *totalResult = [NSMutableArray new];

  [keys enumerateObjectsUsingBlock:^(NSArray *obj, NSUInteger idx, BOOL *stop) {

      [obj enumerateObjectsUsingBlock:^(id obj2, NSUInteger idx2, BOOL *stop2) {
          NSArray *filtered = [totalResult filteredArrayUsingPredicate:[NSPredicate predicateWithFormat:@"(id == %@)", obj2[ @"id" ]]];
          if ( !filtered.count ) {
              [totalResult addObject:obj2];
          }
      }];

  }];

  NSLog(@"");

Answer 2

合并两个词典的键（由dict.allKeys检索的两个单独的数组），并创建一个新词典。但是如果你在allKeys中获得类似的密钥，你必须创建自己的系统来挑选一两个（或者如果你有两个以上的词典）更多的尊重值。

NSDictionary *dict1=[NSDictionary dictionary];
NSDictionary *dict2=[NSDictionary dictionary];

NSArray *arr1=[dict1 allKeys];
NSArray *arr2=[dict2 allKeys];

NSArray *allKeysArray=[arr1 arrayByAddingObjectsFromArray:arr2];

NSMutableDictionary *newDict=[[NSMutableDictionary alloc]init];

for (NSString *aKey in allKeysArray)
{
    BOOL bothArrayContainSameKey=NO;
    if ([arr1 containsObject:aKey] && [arr2 containsObject:aKey])
    {
        bothArrayContainSameKey=YES;
        //needs special treatment
    }
    else if ([arr1 containsObject:aKey])
    {
        [newDict setObject:dict1[aKey] forKey:aKey];
    }
    else if ([arr2 containsObject:aKey])
    {
        [newDict setObject:dict2[aKey] forKey:aKey];
    }
}

合并两个NSDictionary而不重复

2 个答案: