Ios NSDictionary数组 – 分组值和键
我有以下结果的数组的NSDictionary
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" ); };
我怎样才能结合的价值观和关键，使我最终只有一个类别如下所示;
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好方法，代码如下
//分配/初始化所有Nssarrays
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); } }
/ /这给错误“无法识别的select器发送到实例”，这里是打印出来
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
你可以创build一个NSMutableDictionary并循环你的数组，使用allKeys将这个键添加到可变字典中。
例如，如果你的数组被称为dictArray ，你可以这样做：
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]]; } }
这个代码首先创build一个字典combinedDictionary ，这将是你的最终字典。 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。 对于你提供的这个数组的字典，其前两个看起来像@[@"Bath"] ，另外两个则是@[@"Birmingam"] 。
然后代码通过这些键循环，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build一个。
然后，它会从数组中添加字典中的所有值，并将新字典设置为combinedDictionary字典。