从 parse.com 检索数据时出错 [英] Error while retrieving data from parse.com

问题描述

您好，我想从我的 parse.com 类中获取一些数据，该类名为Tags"，该类中有两个 3 列objectID"、username"和tagtext".我想通过 ID 和后记读取记录查找我想将用户名"和标签文本"保存为两个字符串.我已经做到了，就像在 parse.com 文档中一样:

Hello I want to get some data from my parse.com class called "Tags" in this class there are two 3 cols "objectID", "username" and "tagtext". I want to read a record finding by ID and afterwords I want to save "useername" and "tagtext" into two strings. I have done it like it is in the parse.com documentation:

@IBAction func readAction(sender: UIButton) {

    var query = PFQuery(className:"Tags")
    query.getObjectInBackgroundWithId("IsRTwW1dHY") {
        (gameScore: PFObject?, error: NSError?) -> Void in
        if error == nil && gameScore != nil {
            println(gameScore)
        } else {
            println(error)
        }
    }

    let username = gameScore["username"] as! String
    let tagtext = gameScore["tagtext"] as! String

    println(username)
    println(tagtext)   

}

我收到一个名为 fatal error:unwrapping an Optional value 时意外发现 nil 的错误，请告诉我我的代码有什么问题.

I get an error called fatal error: unexpectedly found nil while unwrapping an Optional value , please tell me what is wrong in my code.

我的班级:

推荐答案

问题在于:

let username = gameScore["username"] as! String
let tagtext = gameScore["tagtext"] as! String

gameScore["username"] 和 gameScore["tagtext"] 可以返回 nil 值，当你说 as!String 你说它会是一个String，它是nil.

gameScore["username"] and gameScore["tagtext"] can return nil values, and when you say as! String you say that it will be a String, and it is nil.

尝试以下操作:

let username = gameScore["username"] as? String
let tagtext = gameScore["tagtext"] as? String

您的错误因此而发生，但您的最终代码应如下所示:

your error is happening because of that, but your final code should look like this:

@IBAction func readAction(sender: UIButton) {

  var query = PFQuery(className:"Tags")
  query.getObjectInBackgroundWithId("f3AXazT9JO") {
    (gameScore: PFObject?, error: NSError?) -> Void in

    let username = gameScore["username"] as? String
    let tagtext = gameScore["tagtext"] as? String

    println(username)
    println(tagtext) 
    if error == nil && gameScore != nil {
       println(gameScore)
    } else {
       println(error)
    }
  }
}

因为 getObjectInBackgroundWithId 是异步的.

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