3

我有以下NSDictionary数组结果

Bath =     {
    Keynsham =         (
        "nsham companies"
    );
};


Bath =     {
    "Midsomer Norton" =         (
        "Keynsham companies"
    );
};


Bath =     {
    "Norton Radstock" =         (
        "Keynsham taxi companies"
    );
};


Birmingham =     {
    "Acock's Green" =         (
        "Acock's Green taxi companies"
    );
};


Birmingham =     {
    "Alcester Lane's End" =         (
        "Alcester Lane's End taxi companies"
    );
};

我如何组合值和键，以便我最终只得到一个类别，如下所示；

Bath =     {
    "Norton Radstock" =         (
        "Keynsham taxi companies"
    );
 "Midsomer Norton" =         (
        "Keynsham companies"
    );

   Keynsham =         (
        "nsham companies"
    );

};

我不确定这是否是解释它的最佳方式，代码如下

//所有Nssarrays分配/初始化

  NSURL *url=[NSURL URLWithString:@"http://y.php"];
        NSData *data= [NSData dataWithContentsOfURL:url];

        NSMutableArray *json = [NSJSONSerialization JSONObjectWithData:data options:kNilOptions error:Nil];
        //instantiate arrays to hold data

        NSMutableDictionary *dictArray=[[NSMutableDictionary alloc]init];
        NSArray *cityName=[[NSArray alloc]init];
        NSArray *townName=[[NSArray alloc]init];
        NSArray *taxis=[[NSArray alloc]init];  

        NSArray *ids=[[NSArray alloc]init];

        for (int i=0; i<json.count; i++)
        {

            //cityName=[[NSMutableArray alloc] initWithCapacity:json.count];

           ids = [[json objectAtIndex:i] objectForKey:@"id"];
           cityName = [[json objectAtIndex:i] objectForKey:@"cityName"];
           townName=[[json objectAtIndex:i] objectForKey:@"townName"];

           taxis=[[json objectAtIndex:i] objectForKey:@"taxis"];



        NSMutableArray  *taxisArray=[[NSMutableArray  alloc] initWithObjects:taxis,nil];
       NSMutableDictionary *towensdict=[[ NSMutableDictionary alloc]  initWithObjectsAndKeys:taxisArray,townName, nil];


       NSMutableDictionary *cities1=[[NSMutableDictionary alloc] initWithObjectsAndKeys:towensdict,cityName, nil];


NSLOG (@"%@", cities1) here, gives me the print out above


            [dictArray addEntriesFromDictionary:cities1 ];



Then I tried Jdodgers solution as follows;
   NSMutableDictionary *combinedDictionary = [[NSMutableDictionary alloc] init];
            for (NSDictionary *currentDictionary in dictArray) {
                NSArray *keys = [currentDictionary allKeys];
                 for (int n=0;n<[keys count];n++) {
                  NSMutableDictionary *dictionaryToAdd = [combinedDictionary valueForKey:[keys objectAtIndex:n]];
                if (!dictionaryToAdd) dictionaryToAdd = [[NSMutableDictionary alloc] init];
               [dictionaryToAdd setValuesForKeysWithDictionary:[currentDictionary valueForKey:[keys objectAtIndex:n]]];
               [combinedDictionary setValue:dictionaryToAdd forKey:[keys objectAtIndex:n]];

                NSLog(@"%@", currentDictionary);
                }
            }

//这给出错误“无法识别的选择器发送到实例”，这是打印输出

combinedDictionary  NSMutableDictionary *   0x000000010012e580
currentDictionary   NSDictionary *const 0x0000000100116460
dictArray   NSMutableDictionary *   0x000000010012e220
[0] key/value pair  
key id  0x0000000100116460
[0] id  
value   id  0x000000010012e440
[0] id  
keys    NSArray *   0x0000000000000000
4

1 回答 1

3

您可以创建一个 NSMutableDictionary 并遍历您的数组，使用allKeys.

例如，如果您的数组被调用dictArray，您可以这样做：

NSMutableDictionary *combinedDictionary = [[NSMutableDictionary alloc] init];
for (NSDictionary *currentDictionary in dictArray) {
    NSArray *keys = [currentDictionary allKeys];
    for (int n=0;n<[keys count];n++) {
        NSMutableDictionary *dictionaryToAdd = [combinedDictionary valueForKey:[keys objectAtIndex:n]];
        if (!dictionaryToAdd) dictionaryToAdd = [[NSMutableDictionary alloc] init];
        [dictionaryToAdd setValuesForKeysWithDictionary:[currentDictionary valueForKey:[keys objectAtIndex:n]]];
        [combinedDictionary setValue:dictionaryToAdd forKey:[keys objectAtIndex:n]];
    }
}

此代码首先创建一个字典combinedDictionary，它将成为您的最终字典。它遍历数组中的所有字典，并为每个字典执行以下操作：

首先，它获取字典中所有键的数组。对于您提供的字典，该数组将类似于@[@"Bath"]前 3 个和@[@"Birmingam"]其他两个。

然后代码循环遍历这些键，并从该键的组合字典中获取已经存在的字典。如果字典不存在，则创建一个。

然后，它从数组中添加字典中的所有值，并将新字典设置为combinedDictionary.

于 2013-09-14T07:08:48.333 回答