Cocoa touch 如何在NSUserDefault中存储CCSprite

我在使用CCSprite子类生物时遇到了一个奇怪的问题

让我们，我的对象是生物

类生物声明-

@interface Creature : CCSprite <NSCoding>{

    int creatureAge;
    NSString *creatureName;
}

问题是当我使用NSUserDefault存储我的生物类对象“生物”时-

- (void)encodeWithCoder:(NSCoder *)encoder {

    [encoder encodeObject:self.creatureName forKey:@"creatureName"];
    [encoder encodeInt:self.creatureAge forKey:@"creatureAge"];
}

然后用计算机解码-

- (id)initWithCoder:(NSCoder *)decoder {
    if((self = [super init])) {

        self.creatureName = [decoder decodeObjectForKey:@"creatureName"];
        self.creatureAge= [decoder decodeIntForKey:@"creatureAge"];
}

然后使用-

NSUserDefaults *defaults = [NSUserDefaults standardUserDefaults];
NSData *myEncodedObject = [NSKeyedArchiver archivedDataWithRootObject:creature];

[defaults setObject:myEncodedObject forKey:@"my_pet"];

而负载—

NSUserDefaults *defaults = [NSUserDefaults standardUserDefaults];
 NSData *myEncodedObject = [defaults objectForKey:@"my_pet"];
 Creature* newcreature = (Creature *)[NSKeyedUnarchiver unarchiveObjectWithData: myEncodedObject];

问题是，当我加载这个时，我得到了以前存储的生物的属性值，但分配给以前生物的图像可能没有复制。因为如果我将新生物添加到任何CCLayer中，它不会显示任何图像，尽管它获得了前一个生物的属性值

---

现在我该怎么做才能得到有图像的新生物？是否需要将图像名称作为一个单独的属性添加？

您也可以简单地存储类型，然后按如下方式执行：

@interface Creature : CCSprite {
    int creatureType;
    int creatureAge;
    NSString *creatureName;
}

+ (id)creatureWithType:(int)type;
- (id)initWithCreatureType:(int)type;
@end

@implementation Creature

+ (id)creatureWithType:(int)type
{
    return [[[[self class] alloc] initWithCreatureType:type] autorelease];
}

- (id)initWithCreatureType:(int)type
{
    self = [super initWithFile:[NSString stringWithFormat:@"ch%i_default.png", type]];
    if (!self) return nil;

    creatureType = type;

    return self;
}

- (id)initWithCoder:(NSCoder *)decoder {
    int type = [decoder decodeIntForKey:@"creatureType"];
    self = [self initWithCreatureType:type];
    if (!self) return nil;

    self.creatureName = [decoder decodeObjectForKey:@"creatureName"];
    self.creatureAge= [decoder decodeIntForKey:@"creatureAge"];

    return self;
}

- (void)encodeWithCoder:(NSCoder *)encoder
{
    [encoder encodeInt:creatureType forKey:@"creatureType"];
    [encoder encodeObject:self.creatureName forKey:@"creatureName"];
    [encoder encodeInt:self.creatureAge forKey:@"creatureAge"];
}
@end

---

您可能还希望通过属性公开

creatureType

。请注意，与其使用

初始化creatureWithType:

名称，不如使用

creatureWithType:

名称“更多Cocoa”，谢谢您的回复。我知道，但类型是不够的，因为它可能需要ch%I\u默认值、ch%I\u悲伤、ch%I\u快乐等，所以需要NSString。然而，这并不是主要的问题，我不清楚是否需要将图像编码为单独的属性。事实上，起初我认为图像将被包装在对象本身中，所以不需要明确地添加它。有人能解释为什么不会发生吗？？？您可以将生成的字符串（如

ch1_default.png

）存储在一个实例变量中，并按照我用int显示的方式对其进行编码/解码，然后在

initWithCoder:

中调用

self=[self initWithFile:decodedString]。如果要存储图像，请使用将
UIImage
转换为
NSData
，以及
[UIImage imageWithData://code>。好的，答案很可能是-需要对属于特定类的所有内容进行编码/解码。所以，如果我想要与对象对应的图像，那么也需要对其进行编码/解码。
@interface Creature : CCSprite {
    int creatureType;
    int creatureAge;
    NSString *creatureName;
}

+ (id)creatureWithType:(int)type;
- (id)initWithCreatureType:(int)type;
@end

@implementation Creature

+ (id)creatureWithType:(int)type
{
    return [[[[self class] alloc] initWithCreatureType:type] autorelease];
}

- (id)initWithCreatureType:(int)type
{
    self = [super initWithFile:[NSString stringWithFormat:@"ch%i_default.png", type]];
    if (!self) return nil;

    creatureType = type;

    return self;
}

- (id)initWithCoder:(NSCoder *)decoder {
    int type = [decoder decodeIntForKey:@"creatureType"];
    self = [self initWithCreatureType:type];
    if (!self) return nil;

    self.creatureName = [decoder decodeObjectForKey:@"creatureName"];
    self.creatureAge= [decoder decodeIntForKey:@"creatureAge"];

    return self;
}

- (void)encodeWithCoder:(NSCoder *)encoder
{
    [encoder encodeInt:creatureType forKey:@"creatureType"];
    [encoder encodeObject:self.creatureName forKey:@"creatureName"];
    [encoder encodeInt:self.creatureAge forKey:@"creatureAge"];
}
@end