ios - 如何使 Http get 请求 URL 由 swift 3 中的管道组成？

Question

我从 swift 3 发出了一个获取请求。那个 url 有管道。没有管道它工作正常但是当我添加管道时代码中断说在展开可选值错误时发现 nil。

这是我的网址let url = URL(string:"https://maps.googleapis.com/maps/api/directions/json?origin=colombo&destination=kandy&waypoints=optimize:true|&key=AIzaSyCNBcQLIVvNwqjkYiLawnYK_AM ")

这是我的代码:

func synchronusGetRequstForExternalAPI(api_url:String, headers:[ String: String]) -> ResultModel {

        let resultModel = ResultModel()

        //create the url with URL
        let url = URL(string:"https://maps.googleapis.com/maps/api/directions/json?origin=colombo&destination=kandy&waypoints=optimize:true|&key=AIzaSyCNBcQLIVvNwqjkYiLawnYK_OK4EQbRM5M")
        //create the URLRequest object using the url object
        var request = URLRequest(url: url!)

        //set headers
        for item in headers {
            request.addValue(item.value, forHTTPHeaderField: item.key)
        }

        let semaphore = DispatchSemaphore(value: 0)
        let task = URLSession.shared.dataTask(with: request as URLRequest) {
            (data, response, error) in


            if(error != nil){
                resultModel.ErrorType = .NO_INT
                resultModel.JsonReslut = JSON.null

            }else{

                if let resp = response as? HTTPURLResponse{
                    if(resp.statusCode == 200){
                        if let jsonResult = JSON(data) as? JSON {

                            resultModel.ErrorType = .NO_ERROR
                            resultModel.JsonReslut = jsonResult
                        }
                    }else{
                        if let jsonResult = JSON(data) as? JSON {

                            resultModel.ErrorType = .SEREVR_ERROR
                            resultModel.JsonReslut = jsonResult
                        }else{
                            resultModel.ErrorType = .SEREVR_ERROR
                            resultModel.JsonReslut = JSON.null
                        }
                    }
                }
            }

            semaphore.signal()
        }
        task.resume()
        _ = semaphore.wait(timeout: DispatchTime.distantFuture)

        return resultModel
    }

谁能告诉我为什么会这样？

Answer 1

| 不是 URL 中允许的字符。尝试对 URL 字符串进行百分比转义

let url = URL(string:"https://maps.googleapis.com/maps/api/directions/json?origin=colombo&destination=kandy&waypoints=optimize:true|&key=AIzaSyCNBcQLIVvNwqjkYiLawnYK_OK4EQbRM5M".stringByAddingPercentEncodingWithAllowedCharacters(NSCharacterSet.URLFragmentAllowedCharacterSet()))