一尘不染

使用POST方法在Swift中进行HTTP请求

swift

我正在尝试在Swift中运行HTTP请求,以将2个参数发布到URL。

例:

链接: www.thisismylink.com/postName.php

参数:

id = 13
name = Jack

最简单的方法是什么?

我什至不想阅读回复。我只想发送该文件,以通过PHP文件对数据库进行更改。


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2020-07-07

共1个答案

一尘不染

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.percentEncoded()

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 percentEncoded() -> Data? {
        return map { key, value in
            let escapedKey = "\(key)".addingPercentEncoding(withAllowedCharacters: .urlQueryValueAllowed) ?? ""
            let escapedValue = "\(value)".addingPercentEncoding(withAllowedCharacters: .urlQueryValueAllowed) ?? ""
            return escapedKey + "=" + escapedValue
        }
        .joined(separator: "&")
        .data(using: .utf8)
    }
}

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
    }()
}

这将检查基本的网络错误以及高级HTTP错误。这也可以正确地对查询的参数进行转义。

请注意,我使用的nameof Jack & Jill来说明的正确x-www-form- urlencoded结果name=Jack%20%26%20Jill,该结果是“百分比编码”的(即,空格替换为%20&值中的替换为%26)。

2020-07-07