使用POST方法在Swift中进行HTTP请求

Question

我正在尝试在Swift中运行HTTP请求,将POST 2参数发送到URL.

例:

链接: www.thisismylink.com/postName.php

PARAMS:

id = 13
name = Jack

最简单的方法是什么？

我甚至不想阅读回复.我只是想发送它来通过PHP文件对我的数据库进行更改.

Answer 1

在Swift 3及更高版本中,您可以:

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 parameters: [String: Any] = [
    "id": 13,
    "name": "Jack & Jill"
]
request.httpBody = parameters.percentEscaped().data(using: .utf8)

let task = URLSession.shared.dataTask(with: request) { data, response, error in
    guard let data = data, 
        let response = response as? HTTPURLResponse, 
        error == nil else {                                              // check for fundamental networking error
        print("error", error ?? "Unknown error")
        return
    }

    guard (200 ... 299) ~= response.statusCode else {                    // check for http errors
        print("statusCode should be 2xx, but is \(response.statusCode)")
        print("response = \(response)")
        return
    }

    let responseString = String(data: data, encoding: .utf8)
    print("responseString = \(responseString)")
}

task.resume()

哪里:

extension Dictionary {
    func percentEscaped() -> String {
        return map { (key, value) in
            let escapedKey = "\(key)".addingPercentEncoding(withAllowedCharacters: .urlQueryValueAllowed) ?? ""
            let escapedValue = "\(value)".addingPercentEncoding(withAllowedCharacters: .urlQueryValueAllowed) ?? ""
            return escapedKey + "=" + escapedValue
        }
        .joined(separator: "&")
    }
}

extension CharacterSet { 
    static let urlQueryValueAllowed: CharacterSet = {
        let generalDelimitersToEncode = ":#[]@" // does not include "?" or "/" due to RFC 3986 - Section 3.4
        let subDelimitersToEncode = "!$&'()*+,;="

        var allowed = CharacterSet.urlQueryAllowed
        allowed.remove(charactersIn: "\(generalDelimitersToEncode)\(subDelimitersToEncode)")
        return allowed
    }()
}

这将检查基本网络错误以及高级HTTP错误.这也正确地百分比转义查询的参数.

注意,我使用的name的Jack & Jill,以示出正确x-www-form-urlencoded的结果name=Jack%20%26%20Jill,这是"百分比编码"(即空间被替换%20并且&在值被替换为%26).

请参阅Swift 2演绎的此答案的上一版本.

Answer 2

Swift 3和swift 4

@IBAction func submitAction(sender: UIButton) {

    //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

对于任何正在寻找一种干净的方式在 Swift 5 中对 POST 请求进行编码的人。

您不需要手动添加百分比编码。使用URLComponents创建一个GET请求URL。然后使用query该 URL 的属性来正确获取转义查询字符串的百分比。

let url = URL(string: "https://example.com")!
var components = URLComponents(url: url, resolvingAgainstBaseURL: false)!

components.queryItems = [
    URLQueryItem(name: "key1", value: "NeedToEscape=And&"),
    URLQueryItem(name: "key2", value: "vålüé")
]

let query = components.url!.query

该query会是一个正确转义字符串：

key1=NeedToEscape%3DAnd%26&key2=v%C3%A5l%C3%BC%C3%A9

现在您可以创建一个请求并将查询用作 HTTPBody：

var request = URLRequest(url: url)
request.httpMethod = "POST"
request.httpBody = Data(query.utf8)

现在您可以发送请求了。

Answer 4

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

此方法填写Google表单中的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 5

let session = URLSession.shared
        let url = "http://...."
        let request = NSMutableURLRequest(url: NSURL(string: url)! as URL)
        request.httpMethod = "POST"
        request.addValue("application/json", forHTTPHeaderField: "Content-Type")
        var params :[String: Any]?
        params = ["Some_ID" : "111", "REQUEST" : "SOME_API_NAME"]
        do{
            request.httpBody = try JSONSerialization.data(withJSONObject: params, options: JSONSerialization.WritingOptions())
            let task = session.dataTask(with: request as URLRequest as URLRequest, completionHandler: {(data, response, error) in
                if let response = response {
                    let nsHTTPResponse = response as! HTTPURLResponse
                    let statusCode = nsHTTPResponse.statusCode
                    print ("status code = \(statusCode)")
                }
                if let error = error {
                    print ("\(error)")
                }
                if let data = data {
                    do{
                        let jsonResponse = try JSONSerialization.jsonObject(with: data, options: JSONSerialization.ReadingOptions())
                        print ("data = \(jsonResponse)")
                    }catch _ {
                        print ("OOps not good JSON formatted response")
                    }
                }
            })
            task.resume()
        }catch _ {
            print ("Oops something happened buddy")
        }

Answer 6

这里的所有答案都使用 JSON 对象。这给 $this->input->post() 我们的 Codeigniter 控制器的方法带来了问题。无法CI_Controller直接读取 JSON。我们使用这个方法来做到这一点，而不需要 JSON

func postRequest() {
    // Create url object
    guard let url = URL(string: yourURL) else {return}

    // Create the session object
    let session = URLSession.shared

    // Create the URLRequest object using the url object
    var request = URLRequest(url: url)

    // Set the request method. Important Do not set any other headers, like Content-Type
    request.httpMethod = "POST" //set http method as POST

    // Set parameters here. Replace with your own.
    let postData = "param1_id=param1_value&param2_id=param2_value".data(using: .utf8)
    request.httpBody = postData

    // Create a task using the session object, to run and return completion handler
    let webTask = session.dataTask(with: request, completionHandler: {data, response, error in
    guard error == nil else {
        print(error?.localizedDescription ?? "Response Error")
        return
    }
    guard let serverData = data else {
        print("server data error")
        return
    }
    do {
        if let requestJson = try JSONSerialization.jsonObject(with: serverData, options: .mutableContainers) as? [String: Any]{
            print("Response: \(requestJson)")
        }
    } catch let responseError {
        print("Serialisation in error in creating response body: \(responseError.localizedDescription)")
        let message = String(bytes: serverData, encoding: .ascii)
        print(message as Any)
    }

    // Run the task
    webTask.resume()
}

现在您的CI_Controller 将能够获取param1并param2使用$this->input->post('param1')$this->input->post('param2')