Question

我第一次使用云代码，并尝试调用以下函数：

       let friendRequest: PFObject = self.friendRequestsToCurrentUser[sender.tag] as PFObject
      let fromUser: PFUser = friendRequest[FriendRequestKeyFrom] as PFUser
      //call the cloud code function that adds the current user to the user who sent the request and pass in the friendRequest id as a parameter
        PFCloud.callFunctionInBackground("addFriendToFriendsRelation", withParameters: ["friendRequest": friendRequest.objectId]) { (object:AnyObject!, error: NSError!) -> Void in
        let friendsRelation: PFRelation = PFUser.currentUser()!.relationForKey("friends")
         friendsRelation.addObject(fromUser)
             self.currentUser.saveInBackgroundWithBlock({ (succeeded: Bool, error: NSError!) -> Void in
                 if succeeded {
                 } else {
            }
        })
    }
}

实施该功能后，我需要添加“！”到参数中的objectId来解开它。但是，这样做会让我产生错误：

无法转换类型'的值（AnyObject！，NSError！） - ＆gt;无效的期望参数类型'PFIdResultsBlock？'

为了调用此功能，我必须更改什么？

Answer 1

尝试使用PFObject?代替AnyObject!。

Answer 2

PFIdResultsBlock对应于以下签名(AnyObject?, NSError?) -> Void，因此请尝试将您的代码更改为：

let friendRequest: PFObject = self.friendRequestsToCurrentUser[sender.tag] as PFObject
let fromUser: PFUser = friendRequest[FriendRequestKeyFrom] as PFUser

//call the cloud code function that adds the current user to the user who sent the request and pass in the friendRequest id as a parameter
PFCloud.callFunctionInBackground("addFriendToFriendsRelation", withParameters: ["friendRequest": friendRequest.objectId]) { (object:AnyObject?, error: NSError?) -> Void in
    let friendsRelation: PFRelation = PFUser.currentUser()!.relationForKey("friends")
    friendsRelation.addObject(fromUser)
    self.currentUser.saveInBackgroundWithBlock({ (succeeded: Bool, error: NSError!) -> Void in
        if succeeded {

        } else {

        }
    })
}

在解析云代码

2 个答案: