带参数的 Swift GET 请求 [英] Swift GET request with parameters
问题描述
我对 swift 很陌生,所以我的代码中可能会有很多错误,但我想要实现的是向带有参数的本地主机服务器发送一个 GET
请求.更重要的是,鉴于我的函数采用两个参数 baseURL:string,params:NSDictionary
,我正试图实现它.我不确定如何将这两者结合到实际的 URLRequest 中?这是我迄今为止尝试过的
I'm very new to swift, so I will probably have a lot of faults in my code but what I'm trying to achieve is send a GET
request to a localhost server with paramters. More so I'm trying to achieve it given my function take two parameters baseURL:string,params:NSDictionary
. I am not sure how to combine those two into the actual URLRequest ? Here is what I have tried so far
func sendRequest(url:String,params:NSDictionary){
let urls: NSURL! = NSURL(string:url)
var request = NSMutableURLRequest(URL:urls)
request.HTTPMethod = "GET"
var data:NSData! = NSKeyedArchiver.archivedDataWithRootObject(params)
request.HTTPBody = data
println(request)
var session = NSURLSession.sharedSession()
var task = session.dataTaskWithRequest(request, completionHandler:loadedData)
task.resume()
}
}
func loadedData(data:NSData!,response:NSURLResponse!,err:NSError!){
if(err != nil){
println(err?.description)
}else{
var jsonResult: NSDictionary = NSJSONSerialization.JSONObjectWithData(data, options: NSJSONReadingOptions.MutableContainers, error: nil) as NSDictionary
println(jsonResult)
}
}
推荐答案
在构建 GET
请求时,请求没有正文,而是所有内容都在 URL 上进行.要构建 URL(并正确地对其进行百分比转义),您还可以使用 URLComponents
.
When building a GET
request, there is no body to the request, but rather everything goes on the URL. To build a URL (and properly percent escaping it), you can also use URLComponents
.
var url = URLComponents(string: "https://www.google.com/search/")!
url.queryItems = [
URLQueryItem(name: "q", value: "War & Peace")
]
唯一的技巧是大多数 Web 服务需要转义 +
字符百分比(因为它们会将其解释为 application/x-www-form-urlencoded
规范).但是 URLComponents
不会百分百转义它.Apple 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,它在 URI 的查询中是允许的,但它在 application/x-www-form-urlencoded
请求中具有特殊含义,并且真的不应该不转义地传递.
The only trick is that most web services need +
character percent escaped (because they'll interpret that as a space character as dictated by the application/x-www-form-urlencoded
specification). But URLComponents
will not percent escape it. Apple contends that +
is a valid character in a query and therefore shouldn't be escaped. Technically, they are correct, that it is allowed in a query of a URI, but it has a special meaning in application/x-www-form-urlencoded
requests and really should not be passed unescaped.
Apple 承认我们必须对 +
字符进行百分比转义,但建议我们手动进行:
Apple acknowledges that we have to percent escaping the +
characters, but advises that we do it manually:
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")
这是一种不雅的解决方法,但它有效,如果您的查询可能包含 +
字符并且您有一个将它们解释为空格的服务器,Apple 会建议这样做.
This is an inelegant work-around, but it works, and is what Apple advises if your queries may include a +
character and you have a server that interprets them as spaces.
因此,将其与您的 sendRequest
例程相结合,您最终会得到如下结果:
So, combining that with your sendRequest
routine, you end up with something like:
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
并返回一个自定义的 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.
Personally, I'd use JSONDecoder
nowadays and return a custom struct
rather than a dictionary, but that's not really relevant here. Hopefully this illustrates the basic idea of how to percent encode the parameters into the URL of a GET request.
请参阅此答案的先前修订版,了解 Swift 2 和手动百分比转义格式.
See previous revision of this answer for Swift 2 and manual percent escaping renditions.
这篇关于带参数的 Swift GET 请求的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!