Question

我以前从未与web服务器接口，我对xml知之甚少，但对http头一无所知。我基本上需要连接到服务器并传递信息（使用Cocoa touch），这是提供的信息：

Request headers: 
POST https://www.example.org/example 
Content-Type: text/xml 
Content-Length: 638 
Authorization: [[developer-key]]

我收到了这个请求体：

<LetterValues> 
<FullName><![CDATA[John Doe]]></FullName>  
</LetterValues>

请求主体不仅仅是这个，我删除了很多只是为了保持简单。

那么，有人能给我指一个教程或正确的方向，告诉我如何传递这些信息吗？

Answer 1

-(void)xmlParsingInBackground
{
NSString xmlString = [NSString stringWithFormat:@"Your XML REquest"];
NSURL * serviceUrl = [NSURL URLWithString:[NSString stringWithString:@"YOUR URL"]];
NSMutableURLRequest * serviceRequest = [NSMutableURLRequest requestWithURL:serviceUrl];
[serviceRequest setValue:@"text/xml" forHTTPHeaderField:@"Content-type"];
[serviceRequest setHTTPMethod:@"POST"];
[serviceRequest setHTTPBody:[xmlString dataUsingEncoding:NSUTF8StringEncoding]];
NSData *responseData;
NSURLResponse * serviceResponse;
NSError * serviceError;
responseData = [NSURLConnection sendSynchronousRequest:serviceRequest returningResponse:&serviceResponse error:&serviceError];
NSString *resp=[[NSString alloc] initWithData:responseData encoding:NSUTF8StringEncoding];
if(responseData != NULL)
{       
    NSXMLParser *xmlParser = [[NSXMLParser alloc] initWithData:responseData];
    [xmlParser setDelegate:self];
    //[xmlParser setDelegate:parser];       
    BOOL success = [xmlParser parse];

    if(success)
    { // DO SOMETHING HERE
            }

}

做上面的事情，并实现以下内容：

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

 - (void)parser:(NSXMLParser *)parser foundCharacters:(NSString *)string  

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

这将对您有所帮助。：）

添加：从以下链接下载代码，告诉您如何实现上述方法尺寸XML。

友情链接