尝试将 NSMutableDictionary 添加到 NSMutableArray

Question

我是 Objective-C 和 iOS 编程的新手，所以要温柔:)

我正在尝试向 nsmutablearray 添加一个 nsmutabledictionary。我成功了，但没有达到我希望的结果。这是我的代码：

NSMutableDictionary *dictionary = [[NSMutableDictionary alloc] init];
NSMutableDictionary *messages = [[NSMutableDictionary alloc] init];
NSMutableArray *array = [[NSMutableArray alloc] init];



[dictionary setValue:@"lat1" forKey:@"lat"];
[dictionary setValue:@"long1" forKey:@"long"];
[dictionary setValue:@"alt1" forKey:@"alt"];


[messages setObject:dictionary forKey:@"messages"];

[array addObject:messages];

[dictionary setValue:@"lat2" forKey:@"lat"];
[dictionary setValue:@"long2" forKey:@"long"];
[dictionary setValue:@"alt2" forKey:@"alt"];

[messages setObject:dictionary forKey:@"messages"];

[array addObject:messages];


NSLog(@"%@",array);
NSLog(@"%lu",(unsigned long)[array count]);

这里是 NSLog 输出：

2014-06-05 10:29:27.377 dicttest[4863:60b] (
    {
    messages =         {
        alt = alt2;
        lat = lat2;
        long = long2;
    };
},
    {
    messages =         {
        alt = alt2;
        lat = lat2;
        long = long2;
    };
}

) 2014-06-05 10:29:27.386 字典测试[4863:60b] 2

这是我希望实现的目标：

2014-06-05 10:29:27.377 dicttest[4863:60b] (
    {
    messages =         {
        alt = alt1;
        lat = lat1;
        long = long1;
    };
},
    {
    messages =         {
        alt = alt2;
        lat = lat2;
        long = long2;
    };
}

) 2014-06-05 10:29:27.386 字典测试[4863:60b] 2

如果我将字典直接添加到数组中（而不是将字典添加到消息中，然后将其添加到数组中），那么我会得到我正在寻找的输出。有人可以准确地向我解释我做错了什么吗？

Answer 1

它看起来像你想要的：

一个数组

At index 0: 
  A dictionary with a single key "messages"
    A dictionary with keys "alt", "lat", and "long"
At index 1:
  A dictionary with a single key "messages"
    A dictionary with keys "alt", "lat", and "long"

第二个数组条目中的数据应该使用相同的键，但数据不同。正如其他人指出的那样，您的错误是使用单个字典“字典”

当您将对象添加到字典或数组等集合时，该集合包含指向该对象的指针，而不是该对象的副本。如果您将同一个对象多次添加到集合中，则您有 2 个指向同一个对象的指针，而不是 2 个唯一对象。

当您将“字典”对象添加到结构中、对其进行更改并再次添加时，您不会获得预期的结果，因为结构中的两个条目都指向一个字典。当您更改值时，这两个地方都会发生变化。

“消息”字典也是如此。您还需要其中的 2 个。

通过添加新词典、dictionary2 和 messages2 来修复您的代码：

NSMutableDictionary *dictionary = [[NSMutableDictionary alloc] init];
NSMutableDictionary *dictionary2 = [[NSMutableDictionary alloc] init];

NSMutableDictionary *messages = [[NSMutableDictionary alloc] init];
NSMutableDictionary *messages2 = [[NSMutableDictionary alloc] init];

NSMutableArray *array = [[NSMutableArray alloc] init];

[dictionary setValue:@"lat1" forKey:@"lat"];
[dictionary setValue:@"long1" forKey:@"long"];
[dictionary setValue:@"alt1" forKey:@"alt"];

[messages setObject:dictionary forKey:@"messages"];

[array addObject:messages];

[dictionary2 setValue:@"lat2" forKey:@"lat"];
[dictionary2 setValue:@"long2" forKey:@"long"];
[dictionary2 setValue:@"alt2" forKey:@"alt"];

[messages2 setObject: dictionary2 forKey:@"messages"];

[array addObject: messages2];


NSLog(@"%@",array);
NSLog(@"%lu",(unsigned long)[array count]);

您也可以考虑使用对象字面量语法，例如：

dictionary[@"lat"] = @"lat1";
dictionary[@"long"] = @"long1";
dictionary[@"alt"] = @"alt1";

messages[@"messages"] = 字典；

如果你不需要整个事情都是可变的，你甚至可以用一行来做所有事情：

NSMutableArray *array = [
  @[
    @{@"messages": @{@"lat": @"lat1", @"long": @"long1", @"alt": @"alt1"}},
    @{@"messages": @{@"lat": @"lat2", @"long": @"long2", @"alt": @"alt2"}}
  ];

或者让它可变：

NSMutableArray *array = [
  @[
    [@{@"messages":
      [@{@"lat": @"lat1", @"long": @"long1", @"alt": @"alt1"} mutableCopy]} mutableCopy],
    [@{@"messages":
      [@{@"lat": @"lat2", @"long": @"long2", @"alt": @"alt2"} mutableCopy]} mutableCopy]
                         ] mutableCopy];

编辑：要动态添加内容，您可以使用如下方法：（假设数组是实例变量）

- (void) addMessageWithLat: (NSString *) latString 
  long: (NSString *) longString
  alt: (NSString *) altString;
{
  NSMutableDictionary *messages = [[NSMutableDictionary alloc] init];
  NSDictonary *contents = 
    [@{@"lat": latString, 
      @"long": longString, 
      @"alt": altString} 
    mutableCopy];
  messages[@"messages"] = contents;
  [array addObject: messages];
}

Answer 2

问题是您正在同一对象引用中添加新值。所以新的价值将取代旧的价值。只需在 [dictionary setValue:@"lat2" forKey:@"lat"]; 之前添加这一行即可；

dictionary = [NSMutableDictionary alloc]init];

和 [messages setObject:dictionary forKey:@"messages"] 的第二个实例之前的这一行；

messages = [[NSMutableDictionary alloc] init];