viewcontroller 中的标签填充速度非常慢答案

【问题标题】：Labels in viewcontroller are populated very slowviewcontroller 中的标签填充速度非常慢
【发布时间】：2017-07-19 19:37:20
【问题描述】：

我的视图控制器中的标签填充速度极慢。

脚本将用户 ID 发送到 php 脚本。脚本在数组中检索与用户 ID 相关的数据，并将其作为 json 响应发送回我的应用程序。然后我的 viewcontroller 中的标签将填充 json 数组（字典）中的相应数据。

“do{}”中的打印语句会立即在控制台中执行，但视图控制器中标签的填充需要很长时间。将 self.lblStreet.text = self.loadedStreet 放在 do{} 之外，但仍在任务中，它给出了同样缓慢的结果。

class CharDetailViewController: UIViewController {

//Outlets
@IBOutlet weak var lblUID: UILabel!
@IBOutlet weak var lblStreet: UILabel!

//URL to our web service
let URL_GET_DETAILS = "http://somesite.com/script.php"

var passedUID: String!
var loadedStreet: String!

override func viewDidLoad() {
    super.viewDidLoad()

    let url = URL(string: URL_GET_DETAILS)
    lblUID.text = passedUID

    let request = NSMutableURLRequest(url: url! as URL)
    request.httpMethod = "POST";
    let sendUID = lblUID.text
    let postParameters = "sendUID="+sendUID!;
    request.httpBody = postParameters.data(using: String.Encoding.utf8)
    let task = URLSession.shared.dataTask(with:request as URLRequest){
        data, response, error in

        if error != nil{
            print("error is \(error)")
            return;
        }

        do{
        let myJSON =  try JSONSerialization.jsonObject(with: data!, options: .mutableContainers) as? NSDictionary
        if let parseJSON = myJSON {
            var uid : String!
            uid = parseJSON["userid"] as! String?

            self.loadedStreet = parseJSON["street"] as! String!     
            self.lblStreet.text = parseJSON["street"] as! String!

            print(uid)
            print(self.loadedStreet)
         }     
    }
    catch {
        print(error)
        }
    }     
    task.resume()
}
override func didReceiveMemoryWarning() {
    super.didReceiveMemoryWarning()
}
}

【问题讨论】：

标签： ios json swift3 uilabel populate

【解决方案1】：

您不应该在主线程以外的任何地方编辑 UI。您可以使用此代码获取主线程以实现 UI 更新。

dispatch_async(dispatch_get_main_queue(), ^{
    // Update UI stuff here
});

【讨论】：

【解决方案2】：

在主线程上更新您的 UI。 URLSession 让您处于后台线程中。 Swift 3 中的相应代码是

DispatchQueue.main.async {
    //your UI code
}

所以你的代码应该看起来更像这样。

class CharDetailViewController: UIViewController {

 //Outlets
@IBOutlet weak var lblUID: UILabel!
@IBOutlet weak var lblStreet: UILabel!

//URL to our web service
let URL_GET_DETAILS = "http://somesite.com/script.php"

var passedUID: String!
var loadedStreet: String!

override func viewDidLoad() {
    super.viewDidLoad()

    let url = URL(string: URL_GET_DETAILS)
    lblUID.text = passedUID

    let request = NSMutableURLRequest(url: url! as URL)
    request.httpMethod = "POST";
    let sendUID = lblUID.text
    let postParameters = "sendUID="+sendUID!;
    request.httpBody = postParameters.data(using: String.Encoding.utf8)
    let task = URLSession.shared.dataTask(with:request as URLRequest){
         data, response, error in

         if error != nil{
             print("error is \(error)")
             return;
         }

         do{
         let myJSON =  try JSONSerialization.jsonObject(with: data!, options: .mutableContainers) as? NSDictionary

         if let parseJSON = myJSON {
             var uid : String!
             uid = parseJSON["userid"] as! String?

            DispatchQueue.main.async {
              //your UI code
              self.loadedStreet = parseJSON["street"] as! String! 
              self.lblStreet.text = parseJSON["street"] as! String!
            }

           print(uid)
           print(self.loadedStreet)
        }     
    }catch {
     print(error)
    }
 }     
    task.resume()
 }

 override func didReceiveMemoryWarning() {
   super.didReceiveMemoryWarning()
 }
}

【讨论】：