Question

此代码在Swift 1.1中运行良好...只是试图找出1.2中的更改以使其不兼容：

@IBAction func load_click(sender: AnyObject) {

    var query = PFQuery(className: "myClass")
    query.getObjectInBackgroundWithId("MPSVivtvJR", block: { (object:PFObject!, error: NSError) -> Void in

        let theName = object["name"] as String
        let theAge = object["age"] as Int?

        println(theName)
        println(theAge)

    })
}

它给了我错误：无法调用＆＃39; GetObjectInBackgroundWithId＆＃39;使用类型＆＃39;的参数列表（String，block：（PFObject！，NSError） - ＆gt; Void）

有什么想法吗？谢谢！

Answer 1

现在使用Swift 1.2，你应该更加小心打开选项。因此，在您拥有PFObject和NSError的闭包内，要么删除感叹号，要么添加问号以使其成为可选项。

然后，更安全地打开您的物体。请尝试如下：

// You can create this in a separate file where you save your models

struct myUser {
    let name: String?
    let age: Int?
}

// Now this in the view controller

@IBAction func load_click(sender: AnyObject) {
    var query = PFQuery(className: "myClass")
    query.getObjectInBackgroundWithId("MPSVivtvJR", block: {
        (object:PFObject!, error: NSError?) -> Void in

        if let thisName = object["name"] as? String{
            if let thisAge = object["age"] as? Int{
                let user = myUser(name: thisName, age: thisAge)
                println(user)
            }
        }

    })
}

Answer 2

我为此苦苦挣扎，但下面的代码对我有用。

    var query = PFQuery(className: "class")
    query.whereKey("user", equalTo: PFUser.currentUser()!)
    query.orderByDescending("createdAt")
    var object = query.getFirstObject()

    if let pfObject = object {
        data.variable = (pfObject["variable"] as? Float)!
    }

Answer 3

接受的答案对我不起作用。这是做了什么工作：

var query = PFQuery(className: "score")
query.getObjectInBackgroundWithId("j5xBfJ9YXu", block: {
    (obj, error)in
    if let score = obj! as? PFObject {
        println(score.objectForKey("name"))
    } else {
        println(error)
    }
})

Parse和Swift 1.2问题

3 个答案: