Question

我正在尝试使用以下alamofire请求命中API，并将标头作为参数传递。但它没有发生

let url = "http://192.168.1.42/api/v1/userConfig/8b5dbf7d9bdd40d29472be44efa9b7d3"
let secondTok: String = String(Token.get())


let bear = "Bearer " + secondTok

let param = ["Content-Type" : "application/json","Accept" : "application/json", "Authorization" : bear]



Alamofire.request(url, method: .get , parameters: param , encoding: JSONEncoding.default).responseJSON { response in

Answer 1

在您的代码中，您正在制作参数。将其设置为标题

let url = "http://192.168.1.42/api/v1/userConfig/8b5dbf7d9bdd40d29472be44efa9b7d3"
let secondTok: String = String(Token.get())
let bear = "Bearer " + secondTok
let headers = ["Content-Type" : "application/json","Accept" : "application/json", "Authorization" : bear]

   Alamofire.request(url, method: .get , headers: param , encoding: JSONEncoding.default).responseJSON { response in

Answer 2

与标题授权一样使用此类

let url = "http://192.168.1.42/api/v1/userConfig/8b5dbf7d9bdd40d29472be44efa9b7d3"
let headers: HTTPHeaders = [
        "Authorization": "Bearer " + secondTok!,
        "Accept": "application/json"
     ]
Alamofire.request(url, method: .post, parameters: "" as? [String : AnyObject], encoding: JSONEncoding.default, headers: headers).responseJSON { response in

Answer 3

Alamofire有接收标头作为参数的方法。使用它而不是将header放在参数内。

public func request(
    _ url: URLConvertible,
    method: HTTPMethod = .get,
    parameters: Parameters? = nil,
    encoding: ParameterEncoding = URLEncoding.default,
    headers: HTTPHeaders? = nil)
    -> DataRequest

Answer 4

使用get方法的Alamofire

>>> print('abcd\b')  
abcd
>>> print('abcd\b ') 
abc

Alamofire得到请求

4 个答案: