使用POST方法在Swift中使用HTTP请求

我试图在Swift中运行一个HTTP请求，POST 2参数到一个URL。

例：

链接： www.thisismylink.com/postName.php

PARAMS：

 id = 13 name = Jack 

什么是最简单的方法来做到这一点？

我甚至不想读回应。 我只是想通过一个PHP文件发送数据库上的更改。

Answer 1

在Swift 3中，你可以：

 let url = URL(string: "http://www.thisismylink.com/postName.php")! var request = URLRequest(url: url) request.setValue("application/x-www-form-urlencoded", forHTTPHeaderField: "Content-Type") request.httpMethod = "POST" let postString = "id=13&name=Jack" request.httpBody = postString.data(using: .utf8) let task = URLSession.shared.dataTask(with: request) { data, response, error in guard let data = data, error == nil else { // check for fundamental networking error print("error=\(error)") return } if let httpStatus = response as? HTTPURLResponse, httpStatus.statusCode != 200 { // check for http errors print("statusCode should be 200, but is \(httpStatus.statusCode)") print("response = \(response)") } let responseString = String(data: data, encoding: .utf8) print("responseString = \(responseString)") } task.resume() 

这将检查基本的networking错误以及高级别的HTTP错误。

请参阅Swift 2演绎的这个答案的前一个修订 。

Answer 2

Swift 3

 @IBAction func submitAction(sender: AnyObject) { //declare parameter as a dictionary which contains string as key and value combination. considering inputs are valid let parameters = ["id": 13, "name": "jack"] //create the url with URL let url = URL(string: "www.thisismylink.com/postName.php")! //change the url //create the session object let session = URLSession.shared //now create the URLRequest object using the url object var request = URLRequest(url: url) request.httpMethod = "POST" //set http method as POST do { request.httpBody = try JSONSerialization.data(withJSONObject: parameters, options: .prettyPrinted) // pass dictionary to nsdata object and set it as request body } catch let error { print(error.localizedDescription) } request.addValue("application/json", forHTTPHeaderField: "Content-Type") request.addValue("application/json", forHTTPHeaderField: "Accept") //create dataTask using the session object to send data to the server let task = session.dataTask(with: request as URLRequest, completionHandler: { data, response, error in guard error == nil else { return } guard let data = data else { return } do { //create json object from data if let json = try JSONSerialization.jsonObject(with: data, options: .mutableContainers) as? [String: Any] { print(json) // handle json... } } catch let error { print(error.localizedDescription) } }) task.resume() } 

Answer 3

下面是我在日志库中使用的方法： https : //github.com/goktugyil/QorumLogs

此方法填充了Google表单内的HTML表单。

  var url = NSURL(string: urlstring) var request = NSMutableURLRequest(URL: url!) request.HTTPMethod = "POST" request.setValue("application/x-www-form-urlencoded; charset=utf-8", forHTTPHeaderField: "Content-Type") request.HTTPBody = postData.dataUsingEncoding(NSUTF8StringEncoding) var connection = NSURLConnection(request: request, delegate: nil, startImmediately: true) 

Answer 4

 @IBAction func btn_LogIn(sender: AnyObject) { let request = NSMutableURLRequest(URL: NSURL(string: "http://demo.hackerkernel.com/ios_api/login.php")!) request.HTTPMethod = "POST" let postString = "email: [e-mail] & password: testtest" request.HTTPBody = postString.dataUsingEncoding(NSUTF8StringEncoding) let task = NSURLSession.sharedSession().dataTaskWithRequest(request){data, response, error in guard error == nil && data != nil else{ print("error") return } if let httpStatus = response as? NSHTTPURLResponse where httpStatus.statusCode != 200{ print("statusCode should be 200, but is \(httpStatus.statusCode)") print("response = \(response)") } let responseString = String(data: data!, encoding: NSUTF8StringEncoding) print("responseString = \(responseString)") } task.resume() } 

Answer 5

Swift 3 SIMPLEST解决scheme：

 var components = URLComponents(string: "https://www.myAwesomeURL.com/") components?.queryItems = [ URLQueryItem(name: "client_id", value: myClientID), URLQueryItem(name: "client_secret", value: myClientSecret), URLQueryItem(name: "someCode", value: myAwesomeCode), ] guard let url = components?.url else { return } var request = URLRequest(url: url) request.httpMethod = "POST" 

而你去那里你有一个URL和查询参数的发布请求 ！