将 rangeOfString 从 Objective-c 翻译成 Swift

Question

我花了几个小时来翻译这段简短的代码。 Objective C :

NSString *urlStr = [request.URL absoluteString];
NSArray *urlParts = [urlStr componentsSeparatedByString:[NSString stringWithFormat:@"%@/", kREDIRECTURI]];

if (urlParts.count > 1)
{
    urlStr = urlParts[1];
    NSRange token = [urlStr rangeOfString:@"#access_token="];

    if (token.location != NSNotFound)
    {
        vc.access_token = [urlStr substringFromIndex:NSMaxRange(token)];
    }
}

我尝试过的：

Swift :

var urlParts : NSArray = urlStr!.componentsSeparatedByString("\(kREDIRECTURI)")
if urlParts.count > 1
{
    urlStr = urlParts[1] as? String
    var token  = urlStr!.rangeOfString("#access_token=", options: NSStringCompareOptions.allZeros) as NSRange!

    if token != nil
    {
         var vc = ViewController()
         urlStr!.substringFromIndex(NSMaxRange(token))!
    }
}

有什么想法吗？

Answer 1

我将假设 access_token 查询参数在末尾并且代码遵循这样的字符串 'http:\...#access_token='，url 以代码在末尾结束。这是一个简单的方法，可以从 url、

中提取代码

func getTokenFromUrl(url: String) -> String? {

    let range1 = url.rangeOfString("#access_token")

    guard let range = range1 else { return nil }

    let startIndex = range.endIndex.successor()
    let rangeOfToken = url.endIndex

    let rangeOfString = startIndex ..< rangeOfToken

   return  url.substringWithRange(rangeOfString)
}

let url1 = ""

let url2 = ""

getTokenFromUrl(url1) // returns 873jasf82jmsa8sd
getTokenFromUrl(url2) // returns nil

对于 Swift 的早期版本，您可以使用 if let patter 来解包可选的，

func getTokenFromUrl(url: String) -> String? {

    let range1 = url.rangeOfString("#access_token")

    if let range = range1 {

        let startIndex = range.endIndex.successor()
        let rangeOfToken = url.endIndex

        let rangeOfString = startIndex ..< rangeOfToken

       return  url.substringWithRange(rangeOfString)
    }
    return nil
}

将 rangeOfString 从 Objective-c 翻译成 Swift

Translate rangeOfString from Objective-c To Swift

objective-c

ios

nsrange

swift