SWIFT:放慢对Web服务的多个呼叫

Question

我正在处理一个将多个请求发送到Web服务的应用程序。随着我的发展进一步发展，我发现Web服务变得不知所措，我需要减慢一些速度。我想将每个请求分别发送到Web服务，然后等到上一个请求完成后才发送下一个请求。这是使用循环来调用Web服务的功能：

func syncronize(){
    for operation in syncOperations{
        switch operation{
            case "listPhone":
            let listRequest = WSListRequest(requestType: operation, searchCriteria: [SearchCriteria(name: "name", value: "%")], returnTags: [])
            _ = HTTPPost(method: "POST", body: listRequest.xml, operation: operation, credentials: creds)
        default:
            let listRequest = WSLListRequest(requestType: operation, searchCriteria: [SearchCriteria(name: "name", value: "%")], returnTags: ["name"])
            _ = HTTPPost(method: "POST", body: listRequest.xml, operation: operation, credentials: creds)
        }
    }
}

HTTPPOST函数看起来像这样：

class HTTPPost: NSObject, URLSessionDelegate {
var componentDebug = false
var user = String()
var password = String()
var server = String()
var port = String()
var body = NSString()
var response = Data()
init(method: String, body: NSString, operation: String, credentials: WSCredential){
    super.init()
    let bodyData = body.data(using: String.Encoding.utf8.rawValue)
    let config = URLSessionConfiguration.default
    let userPasswordString = NSString(format: "%@:%@", credentials.userName, credentials.password)
    let userPasswordData = userPasswordString.data(using: String.Encoding.utf8.rawValue)
    let base64EncodedCredential = userPasswordData!.base64EncodedString(options: NSData.Base64EncodingOptions.lineLength64Characters)
    let authString = "Basic (base64EncodedCredential)"
    config.httpAdditionalHeaders = ["Authorization" : authString, "Content-Type" : "text/xml;charset=UTF-8"]
    config.timeoutIntervalForRequest = 10.0
    // create the user request
    let urlString = NSString(format: "https://%@:%@/ws/", credentials.server, credentials.port)
    let url = URL(string: urlString as String)
    var request = URLRequest(url: url!)
    request.httpMethod = method
    request.httpBody = bodyData
    request.setValue("Basic (base64EncodedCredential)", forHTTPHeaderField: "Authorization")
    let session = Foundation.URLSession(configuration: config, delegate: self, delegateQueue:OperationQueue.main)
    _ = session.dataTask(with: request, completionHandler: { (data, response, error) in
        let responseParser = XMLParser(data: data!)
        let responseParserDelegate = XMLResponseParser(operation: operation)
        responseParser.delegate = responseParserDelegate
        responseParser.parse()
        // DEBUGGING OPTIONS
        //print(response)
        //print(NSString(data: data!, encoding: NSUTF8StringEncoding))
        DispatchQueue.main.async(execute: {
            self.response = data!
        })

    }).resume()
}

func urlSession(_ session: URLSession, didReceive challenge: URLAuthenticationChallenge, completionHandler: @escaping (URLSession.AuthChallengeDisposition, URLCredential?) -> Void) {
    completionHandler(Foundation.URLSession.AuthChallengeDisposition.useCredential, URLCredential(trust: challenge.protectionSpace.serverTrust!))
}
}

在异型编程方面，我是新手，可以使用一些帮助。预先感谢。

Answer 1

使用信号量，以使一个任务直到上一个任务完成。这是一个演示

// The semaphore value is like the size of a token pool. After you've taken
// all the tokens in the pool, you must wait until a task returns its token
// back to the pool. Here we only have 1 token (1 request at a time)
let semaphore = DispatchSemaphore(value: 1)
// task1 is a request that will take at least 5 seconds to respond
let task1 = URLSession.shared.dataTask(with: URL(string: "https://httpbin.org/delay/5")!) { data, response, error in
    print("Task1 is done")
    semaphore.signal()  // release the token
}
// task2 is a faster request
let task2 = URLSession.shared.dataTask(with: URL(string: "https://httpbin.org")!) { data, response, error in
    print("Task2 is done")
    semaphore.signal()  // release the token
}
// Never wait on your main queue, always do that in the background
DispatchQueue.global(qos: .background).async {
    semaphore.wait() // take a token, wait if needed.
                     // There will never be a wait here, but included for consistency
    print("starting task 1")
    task1.resume()
    semaphore.wait() // take a token, wait if needed
    print("starting task 2")
    task2.resume()
}

使用信号量，输出是您期望的：

starting task 1
Task1 is done
starting task 2
Task2 is done

取出2条semaphore.wait()行，可以看到这两个请求将如何同时发送：

starting task 1
starting task 2
Task2 is done
Task1 is done