Ios 获取要保存的likeCount时出错

我正在尝试用likeCount更新ios中的对象。下面是我当前的代码和错误

NSNumber *likeCount = [self.currentItem.pfObj valueForKey:@"likeCount"];
  NSLog(@"Initial number of likes --->  %@", likeCount);
  likeCount = [NSNumber numberWithInt:[likeCount intValue] + 1];
  NSLog(@"New number of likes --->  %@", likeCount);

  PFQuery *query = [PFQuery queryWithClassName:@"MainItem"];
  [query getObjectInBackgroundWithId:self.currentItem.likeCount
                block:^(PFObject *upLikeCount, NSError *error) {
  NSLog(@"Post query number of likes --->  %@", likeCount);
  upLikeCount = likeCount;
  NSLog(@"New count of likes --->  %@", upLikeCount);
  [upLikeCount saveInBackground];
       }];

like计数在日志中按预期增加，但我无法将其保存到对象。我最初是用count创建一个新对象。现在（上面的代码）我正在查询项，但在xCode和

在日志中：

Initial number of likes --->  1
New number of likes --->  2
-[__NSCFNumber length]: unrecognized selector sent to instance 0x17d12345
*** Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: '-[__NSCFNumber length]: unrecognized selector sent to instance 0x17d12345'

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感谢您的帮助

您需要将值指定给对象中的字段，而不是对象本身

此外，由于您已经拥有该对象，因此无需先获取它-您只需保存它即可-

NSNumber *likeCount = [self.currentItem.pfObj valueForKey:@"likeCount"];
NSLog(@"Initial number of likes --->  %@", likeCount);
likeCount = [NSNumber numberWithInt:[likeCount intValue] + 1];
NSLog(@"New number of likes --->  %@", likeCount);

self.currentItem.pfObj[@"likeCount"]=likeCount;

[self.currentItem.pfObj saveInBackground];

您可以使用

PFObject

方法

incrementKey

进一步简化代码

[self.currentItem.pfObj incrementKey:@"likeCount"];
[self.currentItem.pfObj saveInBackground];

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谢谢你，保罗！这正是我需要的。我被别人甩了