在构建GET
个请求时,请求没有正文,而是所有内容都放在URL上.要构建URL(并对其进行适当的百分比转义),还可以使用URLComponents
.
var url = URLComponents(string: "https://www.google.com/search/")!
url.queryItems = [
URLQueryItem(name: "q", value: "War & Peace")
]
唯一的诀窍是大多数web服务都需要+
个字符的转义(因为它们会将其解释为application/x-www-form-urlencoded
specification指定的空格字符).但URLComponents
%的人无法逃脱.苹果认为+
是查询中的有效字符,因此不应转义.从技术上讲,它们是正确的,在URI的查询中允许这样做,但它在application/x-www-form-urlencoded
个请求中有特殊的意义,并且不应该在未经转移的情况下传递.
苹果承认,我们必须对这+
个字符进行百分制转义,但建议我们手动操作:
var url = URLComponents(string: "https://www.wolframalpha.com/input/")!
url.queryItems = [
URLQueryItem(name: "i", value: "1+2")
]
url.percentEncodedQuery = url.percentEncodedQuery?.replacingOccurrences(of: "+", with: "%2B")
这是一个不优雅的变通办法,但它是有效的,如果你的查询可能包括+
个字符,并且你的服务器将它们解释为空格,苹果就会建议你这样做.
因此,将其与你的sendRequest
个常规动作结合起来,你最终会得到如下结果:
func sendRequest(_ url: String, parameters: [String: String], completion: @escaping ([String: Any]?, Error?) -> Void) {
var components = URLComponents(string: url)!
components.queryItems = parameters.map { (key, value) in
URLQueryItem(name: key, value: value)
}
components.percentEncodedQuery = components.percentEncodedQuery?.replacingOccurrences(of: "+", with: "%2B")
let request = URLRequest(url: components.url!)
let task = URLSession.shared.dataTask(with: request) { data, response, error in
guard
let data = data, // is there data
let response = response as? HTTPURLResponse, // is there HTTP response
200 ..< 300 ~= response.statusCode, // is statusCode 2XX
error == nil // was there no error
else {
completion(nil, error)
return
}
let responseObject = (try? JSONSerialization.jsonObject(with: data)) as? [String: Any]
completion(responseObject, nil)
}
task.resume()
}
你会这样称呼它:
sendRequest("someurl", parameters: ["foo": "bar"]) { responseObject, error in
guard let responseObject = responseObject, error == nil else {
print(error ?? "Unknown error")
return
}
// use `responseObject` here
}
就我个人而言,我现在会使用JSONDecoder
,并返回自定义struct
而不是字典,但这在这里并不重要.希望这说明了如何将参数百分比编码到GET请求的URL中的基本思想.
有关SWIFT 2和手动转义百分比格式副本,请参见previous revision of this answer.