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parse.com类中的标签"中获取一些数据,该类中有两个3个cols"objectID","username"和"tagtext".我想读取通过ID和后缀查找的记录,我想将用户名"和标记文字"保存到两个字符串中.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: unexpectedly found nil while unwrapping an Optional value
的错误,请告诉我我的代码出了什么问题.
I get an error called fatal error: unexpectedly found nil while unwrapping an Optional value
, please tell me what is wrong in my code.
我的课:
推荐答案
问题是:
let username = gameScore["username"] as! String
let tagtext = gameScore["tagtext"] as! String
gameScore["username"]
和gameScore["tagtext"]
可以返回nil值,当您说as! String
时,您说它将是一个String,并且它是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.
尝试类似的东西:
let username = gameScore["username"] as? String
let tagtext = gameScore["tagtext"] as? String
您的错误因此而发生,但是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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