[英]Swift function to return multiple types
我正在尝试构建一个名为makePostBuilder
的实用程序 function ,它看起来像这样。
fileprivate func makePostRequest(apiUrl: String, params: [String: String]) -> URLRequest? {
// build url
let urlString = "\(apiUrl)"
guard let serviceUrl = URL(string: urlString) else { return nil }
// build url request
var request = URLRequest(url: serviceUrl)
request.httpMethod = "POST"
request.setValue("Application/json", forHTTPHeaderField: "Content-Type")
//set http body for url request with incoming params
guard let httpBody = try? JSONSerialization.data(withJSONObject: params, options: []) else {
return nil
}
request.httpBody = httpBody
return request
}
这个 function 的返回类型显然是不正确的。 我希望它返回一个URLRequest
实例或一个Error
实例。 错误实例主要是因为可选的展开可以返回有价值的消息,而不仅仅是nil
(根据当前的实现。
我在思考typealias
,但我不确定这是否是正确的方法。
//not sure if this is right
typealias CustomRequestType = (URLRequest, Error)
在适当的类型定义结束时,我希望 function 看起来像这样
fileprivate func makePostRequest(apiUrl: String, params: [String:String]) -> CustomType {
let urlString = apiUrl
guard let serviceUrl = URL(string: urlString) else { return //error based on customtype? }
// build url request
var request = URLRequest(url: serviceUrl)
request.httpMethod = "POST"
request.setValue("Application/json", forHTTPHeaderField: "Content-Type")
//set http body for url request with incoming params
guard let httpBody = try? JSONSerialization.data(withJSONObject: params, options: []) else {
return //error type
}
//return success type
request.httpBody = httpBody
return request
}
我想我已经接近了,但还没有完全达到。 如果社区可以向我指出一些文档,我也很高兴!
更新:可能的解决方案?
//does this seem plausible?
enum DCError: String, Error {
case invalidUrl = "the url seems to be invalid"
}
typealias DCUrlRequestType = Result<URLRequest, Error>
fileprivate func makePostRequest(apiUrl: String, params: Dictionary<String, String>) -> DCUrlRequestType {
let urlString = apiUrl
guard let serviceUrl = URL(string: urlString) else {
return DCUrlRequestType.failure(DCError.invalidUrl)
}
var request = URLRequest(url: serviceUrl)
request.httpMethod = "POST"
request.setValue("Application/json", forHTTPHeaderField: "Content-Type")
return DCUrlRequestType.success(request)
}
有一个内置的enum Result
可以用作返回类型。 在无法创建URL
的情况下,您可以使用URLError(code: .badURL)
(由 Foundation 提供)。 因此:
fileprivate func makePostRequest(apiUrl: String, params: [String: String]) -> Result<URLRequest, Error> {
let urlString = "\(apiUrl)"
guard let serviceUrl = URL(string: urlString) else {
return .failure(URLError(.badURL))
}
var request = URLRequest(url: serviceUrl)
request.httpMethod = "POST"
request.setValue("application/json", forHTTPHeaderField: "Content-Type")
do {
request.httpBody = try JSONSerialization.data(withJSONObject: params, options: [])
} catch {
return .failure(error)
}
return .success(request)
}
编写此 function 的自然方法是声明它throws
,如下所示:
fileprivate func makePostRequest(apiUrl: String, params: [String: String]) throws -> URLRequest? {
let urlString = "\(apiUrl)"
guard let serviceUrl = URL(string: urlString) else {
throw URLError(.badURL)
}
var request = URLRequest(url: serviceUrl)
request.httpMethod = "POST"
request.setValue("application/json", forHTTPHeaderField: "Content-Type")
request.httpBody = try JSONSerialization.data(withJSONObject: params, options: [])
return request
}
然后让调用者在想要处理错误时使用do/catch
,而不是让调用者切换Result
案例。
如果调用者真的想要一个Result
,她可以使用Result(catching:)
初始化器,它(带有尾随闭包语法)如下所示:
let requestResult = Result { try makePostRequest(apiUrl: urlString, params: [:]) }
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