从字符串创建 URL带编码

如何解决从字符串创建 URL带编码

我正在尝试从字符串创建 URL，如下所示

let searchParam = "?name=movies&Genre=#Action"

func searchMovies(searchString: String) {
    let encodedString = searchParam.encodeSearchString()
    let urlString = "https://search.movies.local/list.html" + encodedString

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

    print("URL: ",url)
}

func encodeSearchString() -> String? {
  let unreserved = "#?=&"
  let allowed = NSMutableCharacterSet.alphanumeric()
  allowed.addCharacters(in: unreserved)
  return addingPercentEncoding(withAllowedCharacters: allowed as CharacterSet)
}

当搜索参数为“?name=movies&Genre=#Action”时它工作正常，但如果搜索参数包含多个#，则 URL 为零。

例如，如果 searchParam 是“?name=#movies&Genre=#Action”

解决方法

问题在于 # 字符是 URL 的 fragment 的分隔符。

构建包含多个组件的 URL 最可靠的方法是 URLComponents 和 URLQueryItem，编码是免费的

func searchMovies(with parameters: [String:String]) {
    var urlComponents = URLComponents(string: "https://search.movies.local/list.html")!
    var queryItems = [URLQueryItem]()
    for (key,value) in parameters {
        queryItems.append(URLQueryItem(name: key,value: value))
    }
    if !queryItems.isEmpty { urlComponents.queryItems = queryItems }
    print("URL: ",urlComponents.url) // https://search.movies.local/list.html?genre=%23action&name=%23movie
}


searchMovies(with: ["name":"#movie","genre":"#action"])

---

或

func searchMovies(by query: String?) {
    var urlComponents = URLComponents(string: "https://search.movies.local/list.html")!
    var queryItems = [URLQueryItem]()
    if let queryString = query {
        let queryPairs = queryString.components(separatedBy: "&")
        for aPair in queryPairs {
            let keyValue = aPair.components(separatedBy: "=")
            if keyValue.count != 2 { continue }
            queryItems.append(URLQueryItem(name: keyValue[0],value: keyValue[1]))
        }
        if !queryItems.isEmpty { urlComponents.queryItems = queryItems }
    }
    print("URL: ",urlComponents.url) // https://search.movies.local/list.html?genre=%23action&name=%23movie
}

searchMovies(by: "name=#movies&Genre=#Action")