iOS - 如何发出SOAP请求并接收关注响应

hp *_*der 5 soap objective-c nsurlconnection

我知道网上有很多关于"如何在iOS中使用SOAP"的东西,但我仍然无法遵循SAOP请求和响应.帮助非常值得赞赏.我使用简单NSURLConnection的请求和响应SOAP Requst

POST ???.asmx HTTP/1.1
Host: ???
Content-Type: text/xml; charset=utf-8
Content-Length: length
SOAPAction: "http://tempuri.org/GetMessages"

<?xml version="1.0" encoding="utf-8"?>
<soap:Envelope xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/">
  <soap:Body>
    <GetMessages xmlns="http://tempuri.org/">
      <GroupName>string</GroupName>
      <Date>dateTime</Date>
    </GetMessages>
  </soap:Body>
</soap:Envelope>
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SOAP响应

HTTP/1.1 200 OK
Content-Type: text/xml; charset=utf-8
Content-Length: length

<?xml version="1.0" encoding="utf-8"?>
<soap:Envelope xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/">
  <soap:Body>
    <GetMessagesResponse xmlns="http://tempuri.org/">
      <GetMessagesResult>xml</GetMessagesResult>
    </GetMessagesResponse>
  </soap:Body>
</soap:Envelope>
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这是我为发送请求而编写的代码..

NSString *soapMessage = [NSString stringWithFormat:
                         @"<?xml version=\"1.0\" encoding=\"utf-8\"?>\n"
                         "<soap:Envelope xmlns:xsi=\"http://www.w3.org/2001/XMLSchema-instance\" xmlns:xsd=\"http://www.w3.org/2001/XMLSchema\" xmlns:soap=\"http://schemas.xmlsoap.org/soap/envelope/\">\n"
                         "<soap:Body>\n"
                         "<GetMessages xmlns=\"http://tempuri.org/\">\n"
                         "<GroupName>%@</GroupName>\n"
                         "<Date>%@</Date>\n"
                         "</GetMessages>\n"
                         "</soap:Body>\n"
                         "</soap:Envelope>\n"
                         , txtfield1.text
                         , textfield2.text
                         ];
NSLog(@"soapMessage: \n%@",soapMessage);

NSURL *url = [NSURL URLWithString:@"???.asmx"];
NSMutableURLRequest *theRequest = [NSMutableURLRequest requestWithURL:url];
NSString *msgLength = [NSString stringWithFormat:@"%d", [soapMessage length]];

[theRequest addValue: @"text/xml; charset=utf-8" forHTTPHeaderField:@"Content-Type"];
[theRequest addValue: @"http://tempuri.org/GetMessages" forHTTPHeaderField:@"SOAPAction"];
[theRequest addValue: msgLength forHTTPHeaderField:@"Content-Length"];
[theRequest setHTTPMethod:@"POST"];
[theRequest setHTTPBody: [soapMessage dataUsingEncoding:NSUTF8StringEncoding]];

NSURLConnection *theConnection = [[NSURLConnection alloc] initWithRequest:theRequest delegate:self];

if(theConnection )
    webData = [[NSMutableData data] retain];
else
    NSLog(@"theConnection is NULL");
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我收到了这个错误 connectionDidFinishLoading

  <soap:Fault>
     <faultcode>soap:Server</faultcode>
     <faultstring>Server was unable to process request. ---&gt; SqlDateTime overflow. Must be between 1/1/1753 12:00:00 AM and 12/31/9999 11:59:59 PM.</faultstring>
     <detail />
  </soap:Fault>
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我提到了这个链接 并 使用了这个链接中的代码并根据我的方便进行了修改以使其工作

我的问题是如何发送请求并收到关注响应

非常感谢提前

Dai*_*jan 2

这个问题似乎与肥皂消息的发送/接收完全无关,而是与您发送的值有关dateTime——它没有正确的格式。

=> 服务器无法处理请求。---> SqlDateTime 溢出。必须介于 1/1/1753 12:00:00 AM 和 12/31/9999 11:59:59 PM 之间。

使用 NSDateFormatter 以请求的格式写入日期字符串