swift - 我怎样才能使用 # 符号而不是 ？在 URLQueryItem 中？

Question

如果我创建这样的 URL:

    guard var urlComponents = URLComponents(url: "https://example.com/something", resolvingAgainstBaseURL: false) else {
        return
    }

    let queryItems = URLQueryItem(name: "token", value: token)
    urlComponents.queryItems = [queryItems]

    guard let urlWithQueryItem = urlComponents.url else {
        return
    }

我希望最终结果类似于 https://example.com/something#token=7as78f6asd678768asd768asd678

而不是默认的https://example.com/something?token=7as78f6asd678768asd768asd678

(我正在寻找比搜索和替换 ? 字符更智能的东西)

谢谢

Answer 1

正如其他人所指出的，URLQueryItem 对象更适合用于查询字符串而不是 anchor /片段。话虽如此，下面的功能可以帮助您完成您的问题。

func constructURL(withURL urlBase: String, andFragmentFromQueryItem queryItem: URLQueryItem) -> URL?{

guard let url = URL(string:urlBase) else {
    return nil
}

guard var urlComponents = URLComponents(url: url, resolvingAgainstBaseURL: false) else {
    return nil
}

let fragmentKey = queryItem.name
let fragmentValue = queryItem.value ?? ""

urlComponents.fragment = "\(fragmentKey)=\(fragmentValue)"

guard let urlWithFragment = urlComponents.url else {
    return nil
}

return urlWithFragment

}

let url = constructURL(withURL:"https://example.com/something", 
andFragmentFromQueryItem: URLQueryItem(name: "token", value: "tokenValue"))
let urlString = url?.absoluteString 
print("\(urlString!)")

这里是代码的工作 Swift fiddle 的链接。 http://play.swiftengine.io/code/8JKI3/2