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问使用Swift 3进行JSON分析
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Stack Overflow用户

提问于 2017-06-08 09:24:05

回答 4查看 296关注 0票数 2

我在我的应用程序中调用一个soap服务。我从服务器获得JSON的响应

{  
  "CustomerName":"Muneer Ahmad/ Ejaz Ahmad (User)  ",
  "IsCorporate":false,
  "ID":48,
  "Name":"Muneer Ahmad/ Ejaz Ahmad (User)  ",
  "GroupID":null,
  "GroupName":null,
  "CustomerID":48
}

我尝试过在线教程的建议，但我无法解析数据。它给我的错误是“字符0周围的无效值”。我的密码是

 let is_URL: String = "http://labs2.unitedtracker.com/WTService.asmx?op=GetTerminalNumber"

    let lobj_Request = NSMutableURLRequest(url: NSURL(string: is_URL)! as URL)
    let session = URLSession.shared

    lobj_Request.httpMethod = "POST"
    lobj_Request.httpBody = is_SoapMessage.data(using: String.Encoding.utf8)
    lobj_Request.addValue("labs2.unitedtracker.com", forHTTPHeaderField: "Host")
    lobj_Request.addValue("text/xml; charset=utf-8", forHTTPHeaderField: "Content-Type")
    lobj_Request.addValue(String(is_SoapMessage.characters.count), forHTTPHeaderField: "Content-Length")
    lobj_Request.addValue("http://tempuri.org/GetTerminalNumber", forHTTPHeaderField: "SOAPAction")

    let task = session.dataTask(with: lobj_Request as URLRequest, completionHandler: {data, response, error -> Void in

        do {
            if let data = data,
                let json = try JSONSerialization.jsonObject(with: data) as? [String: Any],
                let customerID = json["CustomerID"] as? [[String: Any]] {
                print(customerID)
            }
        } catch {
            print("Error deserializing JSON: \(error)")
        }

        if error != nil
        {
            print("Error: " + String(describing: error))
        }
    })
    task.resume()

有人能告诉我我在做什么不对吗？应该如何解析Soap服务的JSON响应。

以下是来自我的服务器的完整响应。

<?xml version="1.0" encoding="utf-8"?><soap:Envelope xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"><soap:Body><GetAuthenticateUserResponse xmlns="http://tempuri.org/"><GetAuthenticateUserResult>[{"CustomerName":"Muneer Ahmad/ Ejaz Ahmad (User)  ","IsCorporate":false,"ID":48,"Name":"Muneer Ahmad/ Ejaz Ahmad (User)  ","GroupID":null,"GroupName":null,"CustomerID":48}]</GetAuthenticateUserResult></GetAuthenticateUserResponse></soap:Body></soap:Envelope>

json

swift

xcode

soap

ios

回答 4

Stack Overflow用户

回答已采纳

发布于 2017-06-08 09:32:58

lobj_Request.addValue("text/xml; charset=utf-8", forHTTPHeaderField: "Content-Type")

如何保证服务器将返回json作为响应？它似乎将返回text/xml作为响应。

关于JSON解析，我建议使用SwiftyJSON。

更多建议

几年前，我曾经使用过soap客户端引擎，它会将wsdl解析为objective-c代码。今天，也许有些工具可以帮助您从Swift生成wsdl代码，以便与soap服务器进行通信。

关于wsdl到快速客户端生成器

https://stackoverflow.com/a/32798709/419348作者有关于wsdltoswift的帖子回答。
http://wsdl2swift.com wsdl2swift url.

票数 1

Stack Overflow用户

发布于 2017-06-08 11:38:59

问题是您从服务器接收XML并尝试将其解析为JSON。您必须首先从XML信封中提取JSON有效负载。以下是整个回应：

<?xml version="1.0" encoding="utf-8"?>
<soap:Envelope xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <soap:Body>
    <GetAuthenticateUserResponse xmlns="http://tempuri.org/">
      <GetAuthenticateUserResult>[{"CustomerName":"Muneer Ahmad/ Ejaz Ahmad (User)  ","IsCorporate":false,"ID":48,"Name":"Muneer Ahmad/ Ejaz Ahmad (User)  ","GroupID":null,"GroupName":null,"CustomerID":48}]</GetAuthenticateUserResult>
    </GetAuthenticateUserResponse>
  </soap:Body>
</soap:Envelope>

您需要将<GetAuthenticateUserResult>标记的内容提取为字符串，只有这样，您才能使用NSJSONSerialization来解析该内容。有点像这样：

class Extractor: NSObject, XMLParserDelegate {

    var payload = ""
    private var recording = false
    let wantedElement = "GetAuthenticateUserResult"

    func parser(_ parser: XMLParser, didStartElement elementName: String, namespaceURI: String?,
        qualifiedName qName: String?, attributes attributeDict: [String : String] = [:]) {
        if elementName == wantedElement {
            recording = true
        }
    }

    func parser(_ parser: XMLParser, foundCharacters string: String) {
        if recording {
            payload += string
        }
    }

    func parser(_ parser: XMLParser, didEndElement elementName: String, namespaceURI: String?, qualifiedName qName: String?) {
        recording = false
    }
}

let response = "<?xml version=\"1.0\" encoding=\"utf-8\"?><soap:Envelope xmlns:soap=\"http://schemas.xmlsoap.org/soap/envelope/\" xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\" xmlns:xsd=\"http://www.w3.org/2001/XMLSchema\"><soap:Body><GetAuthenticateUserResponse xmlns=\"http://tempuri.org/\"><GetAuthenticateUserResult>[{\"CustomerName\":\"Muneer Ahmad/ Ejaz Ahmad (User)\",\"IsCorporate\":false,\"ID\":48,\"Name\":\"Muneer Ahmad/ Ejaz Ahmad (User)  \",\"GroupID\":null,\"GroupName\":null,\"CustomerID\":48}]</GetAuthenticateUserResult></GetAuthenticateUserResponse></soap:Body></soap:Envelope>"
let parser = XMLParser(data: response.data(using: .utf8)!)
let extractor = Extractor()
parser.delegate = extractor
parser.parse()
let payload = extractor.payload
let parsedPayload = try? JSONSerialization.jsonObject(with: payload.data(using: .utf8)!, options: [])

这给了你：

{
    CustomerID = 48;
    CustomerName = "Muneer Ahmad/ Ejaz Ahmad (User)";
    GroupID = "<null>";
    GroupName = "<null>";
    ID = 48;
    IsCorporate = 0;
    Name = "Muneer Ahmad/ Ejaz Ahmad (User)  ";
}

XML提取代码有点愚蠢，因为我不知道有什么更好的内置XML解析器。

票数 1

Stack Overflow用户

发布于 2017-06-08 09:38:58

你的代码：

let customerID = json["CustomerID"] as? [[String: Any]] {
    print(customerID)
}

将customerID视为字典而不是字符串。

它应该是：

let customerID = json["CustomerID"] as? String {
    print(customerID)
}

您应该使用另一种网络请求和JSON解析器的方式，比如阿拉莫火与AlamofireObjectMapper。

票数 0

页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持

原文链接：

https://stackoverflow.com/questions/44431683

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