如何在 swift 4 中获取 ISO8601 格式？

Question

我正在使用亚马逊网络服务进行商品搜索。我可以获得时间戳值并转换为 iso 格式。但在控制台输出中显示 Timestamp 无效。原因：必须是ISO8601格式

这是我的完整控制台输出值：状态码为：可选（400） XML: . InvalidParameterValue参数时间戳的值 2018-06-15T18%3A43%3A52Z 无效。原因：必须是ISO8601格式。66a698a6-a967-4b34-b15a-94780f9287ce

这是我试过的男女同校：

    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
}

}

Answer 1

错误是：

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 : 百分比的转义（转换为 %3A），因为 2018-06-15T18:43:52Z 在 ISO8601 格式中似乎有效，而 2018-06-15T18%3A43%3A52Z 无效。

您在 parameters:

中添加了逃逸百分比

所以：

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

=>

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

另外，请注意 iOS 中有一个 iOS10+ 提供的 ISO8601DateFormatter 可以避免您自己编写格式。

Answer 2

尝试

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
}

如何在 swift 4 中获取 ISO8601 格式？

how to get ISO8601 format in swift 4?

xcode

amazon-web-services

ios

swift

swift3