ios - 如何以编程方式将开关 Controller 的参数快速传递到服务器？

Question

let url = NSURL(string: urlString)

let theRequest = NSMutableURLRequest(URL: url!)

theRequest.HTTPMethod = "POST"

let parameters = ["userId":userId.text!,"status":"offline"] as Dictionary<String,String>


var err:NSError!

do{

    theRequest.HTTPBody = try NSJSONSerialization.dataWithJSONObject(parameters, options: [])
}

catch let error as NSError
{
    err = error
    theRequest.HTTPBody = nil
}

theRequest.addValue("application/json", forHTTPHeaderField: "Content-Type")
theRequest.addValue("application/json", forHTTPHeaderField: "Accept")

如何将开关控制参数传递给服务器？

Answer 1

要传递 bool 值，您需要将其作为另一个参数传递到函数中。例如，现在您有:

let url = NSURL(string: urlString)

将其更改为:

 let url = NSURL(string: urlString, value1: Bool)

然后，当您从 View Controller 调用此 post 函数时，您需要获取与开关值对应的 bool 值的当前状态。你可以这样做:

let value = yourBool.boolValue()

然后，在您的 post 函数中，执行如下操作:

  let url = NSURL(string: urlString, value1: Bool)
  var status = String()
  if value1 == true {
  status = "online"
  }
  else {

  status = "offline"

  }

 let theRequest = NSMutableURLRequest(URL: url!)

 let parameters = ["userId":userId.text!,"status":status]