我有这个我希望在
Swift 3中使用的json数据.我正在学习Swift并构建一个非常基本的应用程序,它显示来自JSON的tableuIView中的项目列表.
{ "expertPainPanels" : [ { "name": "User A","organization": "Company A" },{ "name": "User B","organization": "Company B" } ] }
我正在尝试使用Swift 3获取此数据.
if (statusCode == 200) { do{ let json = try? JSONSerialization.jsonObject(with: data!,options:.allowFragments) // [[String:AnyObject]] /* If I do this: let json = try? JSONSerialization.jsonObject(with: data!,options:.allowFragments) as! [String:Any] if let experts = json?["expertPainPanels"] as! [String: Any] { I get "Initializer for conditional binding must have Optional type,not '[String: Any]'" */ // Type 'Any' has no subscript members. if let experts = json["expertPainPanels"] as? [String: AnyObject] { for expert in experts { let name = expert["name"] as? String let organization = expert["organization"] as? String let expertPainPanel = ExpertPainPanel(name: name,organization: organization)! self.expertPainPanels += [expertPainPanel] self.tableView.reloadData() self.removeLoadingScreen() } } }catch { print("Error with Json: \(error)") } }
它在Swift 2中运行良好.我更新了Swift 3,它破坏了代码.我读过几篇SO,但我仍然很难理解它.我应用了一些建议,包括JSON Parsing in Swift 3,但我仍然无法修复我得到的错误.
从Swift 3开始,你需要尽早进行演员表演.
这一行:
let json = try? JSONSerialization.jsonObject(with: data!,options:.allowFragments)
应该成为这样的:
let json = try JSONSerialization.jsonObject(with: data!,options:.allowFragments) as? [String : AnyObject]
这是因为JSONSerialization现在返回Any,它没有为[]运算符实现变体.确保您安全地展开演员,并采取常见措施,以确保您不会崩溃你的程序.
编辑:您的代码应该或多或少看起来像这样.
let data = Data() let json = try JSONSerialization.jsonObject(with: data,options:.allowFragments) as! [String : AnyObject] if let experts = json["expertPainPanels"] as? [String: AnyObject] {