Question

我有这个对象：

@interface EmailToSend : NSObject

@property (copy, nonatomic) NSString *Subject;
@property (copy, nonatomic) NSString *Body;
@property (strong, nonatomic) NSArray *Cc;
@property (strong, nonatomic) NSArray *Bcc;
@property (strong, nonatomic) NSArray *To;
@property (strong, nonatomic) EmailAddress *From;
@property (copy, nonatomic) NSString *Username;
@property (copy, nonatomic) NSString *Password;

@end

Bcc, To, From as Array EmailAdress

@interface EmailAddress : NSObject

@property (nonatomic, assign) int Id;
@property (copy, nonatomic) NSString *address;
@property (copy, nonatomic) NSString *displayName;

@end

我在iOS中使用JSON框架来解析对象EmailToSend：

 NSData *jsonData = [NSJSONSerialization dataWithJSONObject:emailToSend options:NSJSONWritingPrettyPrinted error:&writeError];
 NSString *jsonString = [[NSString alloc] initWithData:jsonData encoding:NSUTF8StringEncoding];

运行项目时，会出现一个错误：

Terminating app due to uncaught exception
'NSInvalidArgumentException', reason: 'Invalid type in JSON write
(EmailToSend)

如何解决？

Answer 1

EmailToSend是您在应用程序中使用的类类型。但是，Cocoa中的内置JSON序列化程序只能使用简单类型，例如NSString NSArray等。如果您希望它工作，则必须使From属性成为字符串

正如Wain在评论中指出的那样，根元素也必须是NSArray或NSDictionary。

在iOS中将Object转换为json

1 个答案: