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，但我试图实现的是发送 GET 请求到一个带有参数的本地主机服务器。更重要的是我试图实现它，因为我的函数有两个参数 baseURL：string，params：NSDictionary 。我不知道如何将这两个结合到实际的URLRequest中？这是我到目前为止所尝试的结果
func sendRequest（url：String，params：NSDictionary）{
let urls ：NSURL！ = NSURL（string：url）
var request = NSMutableURLRequest（URL：url）
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数据，选项：NSJSONReadingOptions.MutableContainers，error：nil）as NSDictionary
println（jsonResult）
$ b code >
在构建 GET 请求时，这个请求没有任何内容，而是一切都在URL上。要构建一个URL（并正确转换百分比），您还可以使用 URLComponents 。
var url = URLComponents（string：https://www.google.com/search/）！
$ b $ url.queryItems = [
URLQueryItem（name：q，value：War& Peace）
]
唯一的技巧是大多数Web服务需要 + 字符百分比转义（因为他们会解释作为 <$ c $所规定的空格字符c> application / x-www-form-urlencoded specification ）。但 URLComponents 不会百分比转义它。 Apple争辩说， + 是查询中的有效字符，因此不应该转义。从技术上讲，它们是正确的，它允许在一个URI的查询中，但它在 application / x-www-form-urlencoded 请求中有特殊含义，并且确实应该苹果公司承认我们必须转换字符数<％c $ c> + ，但建议我们做它手动：
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）
是一个不起眼的解决方法，但它是有效的，并且是Apple建议的，如果您的查询可能包含一个 + 字符并且您有一个服务器将它们解释为空格。 p>
因此，将它和你的 sendRequest 例程结合起来，就可以得到如下结果：
func sendRequest（_ url：String，parameters：[String：String]，completion：@escaping（[String：Any]？，Error？） - >无效）{
var components = URLComponents（string：url）！
components.queryItems = parameters.map {（key，value）in
URLQueryItem（name：key，value：value）
}
components.percentEncodedQuery = components.percentEncodedQuery ?.用以下代码替换事件：（+，用：％2B）
let request = URLRequest（url：components.url！）
let task = URLSession.shared.dataTask请求）{数据，响应，错误
guard让data = data，//是否有数据
let response = response as？ HTTPURLResponse，//有HTTP响应
（200 ..< 300）〜= response.statusCode，//是statusCode 2XX
错误== nil else {//在那里没有错误，否则。 ..
完成（无，错误）
返回
}
让responseObject =（try？JSONSerialization.jsonObject（with：data））as？ [String：Any]
completion（responseObject，nil）
}
task.resume（）
}
你可以这样称呼它：
sendRequest（someurl ，参数：[foo：bar]）{responseObject，错误
guard let responseObject = responseObject，error == nil else {
print（error ??Unknown error）
return
}
//在这里使用`responseObject`
}
就我个人而言，现在我会使用 JSONDecoder ，并返回一个自定义的 struct 而不是字典，但这不是真的有关。希望这可以说明如何将参数编码到GET请求的URL中的基本思想。
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)
}
}
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")
]
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 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")
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.
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 else { // was there no error, otherwise ...
completion(nil, error)
return
}
let responseObject = (try? JSONSerialization.jsonObject(with: data)) as? [String: Any]
completion(responseObject, nil)
}
task.resume()
}
And you'd call it like:
sendRequest("someurl", parameters: ["foo": "bar"]) { responseObject, error in
guard let responseObject = responseObject, error == nil else {
print(error ?? "Unknown error")
return
}
// use `responseObject` here
}
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.
See previous revision of this answer for Swift 2 and manual percent escaping renditions.
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