ios - 获取要保存的 likeCount 时出错

Question

我正在尝试使用 likeCount 更新 ios 中的对象。以下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];
       }];

点赞数按日志中的预期增加，但我无法将其保存到对象中。我最初是用计数创建一个新对象。现在(上面的代码我正在查询项目但在 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'

感谢您的帮助

Answer 1

您需要将值分配给对象中的字段，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 - 您可以直接保存它 -

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 进一步简化您的代码

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