iphone post请求头

Question

我将开发基于REST API的web服务。这是我的第一个web服务项目。

我有一个关于网络服务的问题--iPhone app。

假设当用户使用iPhone应用程序登录时，他的应用程序将获得身份验证令牌。

使用此令牌，他/她将被授权使用app的进一步操作。我计划关于iPhone应用程序需要发送侦听器与post/get请求方法在牧羊人的身份验证密钥。

请求看起来像: www.mydomaoin.com

但是为了在服务器上授权，我需要将特殊的参数传递到请求的头部：

Headers: X-Testing-Auth-Secret: kI7wGju76kjhJHGklk76
Headers: Content-Type : application/json

它将用于控制器(zend框架)

public function preDispatch()  
{
    $request = new Zend_Controller_Request_Http();
    $key = $request->getHeader('X-Testing-Auth-Secret');
    $type = $request->getHeader('Content-Type');
}

我不知道有没有能发送报头的iPhone。和android的情况一样。

我是一个php开发者。

请帮帮我

Answer 1

您可以使用一些委托方法。

-(void) parser:(NSXMLParser *) parser
didStartElement:(NSString *) elementName``
  namespaceURI:(NSString *) namespaceURI
 qualifiedName:(NSString *) qName
   attributes:(NSDictionary *) attributeDict {

    if( [elementName isEqualToString:@"AddGeneralInformationResult"])
   {
    if (!soapResults)
    {
        soapResults = [[NSMutableString alloc] init];
    }
    elementFound = YES;
   }


    }

  -(void)parser:(NSXMLParser *) parser foundCharacters:(NSString *)string
   {
if (elementFound)
{
    [soapResults appendString: string];
}
}

-(void)parser:(NSXMLParser *)parser
 didEndElement:(NSString *)elementName
  namespaceURI:(NSString *)namespaceURI
  qualifiedName:(NSString *)qName
 {

 if ([elementName isEqualToString:@"AddGeneralInformationResult"])
 {
    elementFound = NO;
    //NSLog(@"AddGeneralInformationResult %@",soapResults);
    if([soapResults isEqualToString:@"true"])
    {


  [objDB updateDatabaseTable:@"GeneralInformation":@"Yes":
    [[arrGenralInfo          objectAtIndex:0] intValue]];
    }
    [soapResults setString:@""];
    elementFound = FALSE;
   }


  }
 -(void) connection:(NSURLConnection *) connection
 didReceiveResponse:(NSURLResponse *) response {
 [webData setLength: 0];

//NSLog(@"receive response");
  }  

 -(void) connection:(NSURLConnection *) connection
 didReceiveData:(NSData *) data {
  [webData appendData:data];
  }

    -(void) connection:(NSURLConnection *) connection
       didFailWithError:(NSError *) error {

    //NSLog(@"Server error");

    }

  -(void) connectionDidFinishLoading:(NSURLConnection *) connection {
     //NSLog(@"DONE. Received Bytes: %d", [webData length]);
      NSString *theXML = [[NSString alloc]
                    initWithBytes: [webData mutableBytes]
                    length:[webData length]
                    encoding:NSUTF8StringEncoding];
//---shows the XML---
//NSLog(@"%@",theXML);

// [activityIndicator stopAnimating];
if (xmlParser)
{

}
xmlParser = [[NSXMLParser alloc] initWithData: webData];
[xmlParser setDelegate: self];
[xmlParser setShouldResolveExternalEntities:YES];
[xmlParser parse];

}

Answer 2

您可以使用以下代码

NSMutableURLRequest *req = [[NSMutableURLRequest alloc]initWithURL:[NSURL URLWithString:url]];
    NSString *post = @"your json";
    //NSLog(@"%@",post);
    NSData *postdata = [NSData dataWithBytes:[post UTF8String] length:[post length] ];
    [req setHTTPMethod:@"POST"];
    [req setValue:@"application/json" forHTTPHeaderField:@"content-type"];
    [req setValue:@"kI7wGju76kjhJHGklk76" forHTTPHeaderField:@"Content-Length"];
    [req setValue:[NSString stringWithFormat:@"%d",[postdata length]] forHTTPHeaderField:@"X-Testing-Auth-Secret"];
    [req setHTTPBody:postdata];

    (void)[NSURLConnection connectionWithRequest:req delegate:self];

Answer 3

是的，你可以从iPhone和安卓手机上发送标题