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如何在Swift 4中获取ISO8601格式?

swift3 amazon-web-services xcode swift ios

PvDev · 技术社区 · 6 年前

以下是我的完整控制台输出值: XML: InvalidParameterValue 参数timestamp的值2018-06-15t18%3a43%3a52z无效。原因:必须采用ISO8601格式。66A698A6-A967-4B34-B15A-94780F9287CE

    public func getSearchItem(searchKeyword: String) -> [String:AnyObject]{
      let timeStamp = getTimestamp()

    let dateFormatter = DateFormatter()
    dateFormatter.timeZone = TimeZone(identifier: "IST")
    let tempLocale = dateFormatter.locale // save locale temporarily
    dateFormatter.locale = Locale(identifier: "en_US_POSIX") // set locale to reliable US_POSIX
    dateFormatter.dateFormat = "yyyy-MM-dd'T'HH:mm:ssZ"
    let date = dateFormatter.date(from: timeStamp.string(from: Date()))!
    dateFormatter.dateFormat = "dd-MM-yyyy HH:mm:ss"
    dateFormatter.locale = tempLocale // reset the locale
    let dateString = dateFormatter.string(from: date)
    print("EXACT_DATE : \(dateString)")

    let url = "http://webservices.amazon.in/onca/xml"
    let parameters: Parameters = ["Service": "AWSECommerceService",
                                  "Operation": "ItemSearch",
                                  "ResponseGroup": "Images,ItemAttributes",
                                  "SearchIndex":"All",
                                  "Keywords": searchKeyword,
                                  "Timestamp": urlEncode(timeStamp.string(from: date)),
                                  "AWSAccessKeyId": urlEncode(ViewController.kAmazonAccessID),
                                  "AssociateTag": urlEncode(ViewController.kAmazonAssociateTag)
                                 ]



    let signedParams = signedParametersForParameters(parameters: parameters as! [String : String])

    Alamofire.request(url, method: .get, parameters: signedParams, encoding: URLEncoding.default)
        .responseString { response in
            print(" - API url: \(String(describing: response.request!))")   // original url request
            var statusCode = response.response?.statusCode

            switch response.result {
            case .success:
                print("status code is: \(String(describing: statusCode))")
                if let string = response.result.value {
                    print("XML: \(string)")
                }
            case .failure(let error):
                statusCode = error._code // statusCode private
                print("status code is: \(String(describing: statusCode))")
                print(error)
            }
    }


    return parameters as [String : AnyObject]
}

func getTimestamp() -> DateFormatter{
    var timestampFormatter = DateFormatter()
    timestampFormatter = DateFormatter()
    timestampFormatter.dateFormat = AWSDateISO8601DateFormat1
    timestampFormatter.timeZone = TimeZone(identifier: "IST")
    timestampFormatter.locale = Locale(identifier: "en_US_POSIX")
    return timestampFormatter
}

}

2 回复 | 直到 6 年前

Larme 6 年前

Optional(400) XML: . InvalidParameterValue Value 2018-06-15T18%3A43%3A52Z for parameter Timestamp is invalid. Reason: Must be in ISO8601 format.

问题是关于 : %3A )因为 2018-06-15T18:43:52Z 2018-06-15T18%3A43%3A52Z 不。

parameters 以下内容:

"Timestamp": urlEncode(timeStamp.string(from: date))

"Timestamp": timeStamp.string(from: date)

ISO8601DateFormatter 可以从IOS10+获得,这样可以避免您自己编写格式。

Supanat Techasothon 6 年前

func convertDateToDateString(date: Date, format: String) -> String {
    let dateFormatter = DateFormatter()
    dateFormatter.locale = Locale(identifier: "en_EN")
    dateFormatter.dateFormat = format
    dateFormatter.calendar = Calendar(identifier: .iso8601)

    let resultDate = dateFormatter.string(from: date)
    return resultDate
}