[NSNull isEqualToString:]

Question

我试图将一个字符串设置为“如果object objectForKey@"display_name”为空，则不显示名称“。它在原木上撞坏了

2011-06-16 10:58:36.251 BV API[15586:ef03] displayNameType is: NSNull
2011-06-16 10:58:36.251 BV API[15586:ef03] +[NSNull isEqualToString:]: unrecognized selector sent to class 0x1228c40
2011-06-16 10:58:36.253 BV API[15586:ef03] *** Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: '+[NSNull isEqualToString:]: unrecognized selector sent to class 0x1228c40'
*** First throw call stack:
(0x1198f1c 0x132b52e 0x119bb2b 0x10f3076 0x10f2bb2 0x116d9 0x112ad 0x8b0e17 0x8b9783 0x8b43ae 0x8b9d20 0xb6312b 0x815c35 0xac2e1e 0x8b7583 0x8b771d 0x8b775d 0xc1b5 0x7ed84c 0x7ed7e2 0x89477a 0x894c4a 0x893ee4 0x813002 0x81320a 0x7f960c 0x7ecd52 0x211b8f6 0x116831a 0x10c3d07 0x10c1e93 0x10c1750 0x10c1671 0x211a0c3 0x211a188 0x7eac29 0x1b29 0x1aa5)
terminate called throwing an exceptionsharedlibrary apply-load-rules all
Current language:  auto; currently objective-c

NSString *displayNameType = (NSString *)[[object objectForKey:@"display_name"] class ];
NSLog(@"displayNameType is: %@", displayNameType);

NSString *displayNameString = [NSString stringWithFormat:@"%@", [object objectForKey:@"display_name"]];
displayNameString = [displayNameString uppercaseString];
if ([displayNameType isEqualToString:@"NSNull"]) {
    NSLog(@"dnt is null");
    NSString *displayNameString = @"No Display Name";
    displayNameString = [displayNameString uppercaseString];
}

Answer 1

强制转换不会更改对象class (类型)。

当您的值为[NSNull null]时，您必须管理这种情况，如下所示：

id displayNameTypeValue = [object objectForKey:@"display_name"];
NSString *displayNameType = @"";
if (displayNameTypeValue != [NSNull null])
   displayNameType = (NSString *)displayNameTypeValue;

Answer 2

我在NSNull上创建了一个类别，非常适合我：

@interface NSNull (string)

-(BOOL) isEqualToString:(NSString *) compare;

@end

@implementation NSNull (string)

-(BOOL) isEqualToString:(NSString *) compare {    

    if ([compare isKindOfClass:[NSNull class]] || !compare) {
        NSLog(@"NSNull isKindOfClass called!");
        return YES;
    }

    return NO;  
}
@end

Answer 3

我将提交我的答案，以澄清一点。问题是NSDictionary中的对象具有NSNull空值(与零和空都略有不同)。这个问题来自于一些库(在我看来是一个特定的JSON解析器)，将一些带空值的键的值设置为NSNull null。为什么？因为有时需要在NSDictionary中区分密钥不存在时的情况和键具有空值时的情况。在NSDictionary中，无法判断，但是JSON结构确实传达了这样的区别。当您得到一个来自库或解析器的变量时，这个值可能是NSNull null。NSNull是一个单例，因此只检查相等性(指针相等)就足够了。例如，我会这样做：

NSString *value = [object objectForKey:@"display_name"];
if (value && value != [NSNull null]){
   // Here I have the value I expect

   // Do something
}else{
   // The value is null ... don't display
}