将XML数据从iphone发布到web服务器
将XML数据从iphone发布到web服务器
提问于 2011-02-07 13:03:28
我想用以下格式发布xml数据：
<?xml version="1.0" encoding="UTF-8"?>
<data>
<email>xyz@domain.com</email>
<password>xyz123</password>
</data>并以以下格式从the服务器接收
<?xml version="1.0" encoding="UTF-8"?>
<user>
<user_id>12</user_id>
</user>现在我正在尝试使用NSUrlConnection和NSMutableRequest，我可以像表单post一样在名称上发布数据，但我只想发布xml数据。我也尝试过使用ASIHTTPRequest。任何代码或链接都是高度精准的。
回答 2
Stack Overflow用户
回答已采纳
发布于 2011-02-07 16:39:14
最后，我使用了来自http://allseeing-i.com/ASIHTTPRequest/的库
并从以下代码发布：
NSURL *url = [NSURL URLWithString:@"http://url to post data"];
ASIFormDataRequest *request = [ASIFormDataRequest requestWithURL:url];
[request addPostValue:@"test name" forKey:@"name"];
[request setDelegate:self];
[request startSynchronous];并使用委托方法接收数据或错误。
- (void)requestFinished:(ASIHTTPRequest *)request
{
// Use when fetching text data
NSString *responseString = [request responseString];
NSLog(@"%@",responseString);
// Use when fetching binary data
NSData *responseData = [request responseData];
}
- (void)requestFailed:(ASIHTTPRequest *)request
{
NSError *error = [request error];
NSLog(@"Error %@",error);
}Stack Overflow用户
发布于 2011-02-07 14:47:20
//prepare request
NSString *urlString = [NSString stringWithFormat:@"http://urlToSend.com"];
NSMutableURLRequest *request = [[[NSMutableURLRequest alloc] init] autorelease];
[request setURL:[NSURL URLWithString:urlString]];
[request setHTTPMethod:@"POST"];
//set headers
NSString *contentType = [NSString stringWithFormat:@"text/xml"];
[request addValue:contentType forHTTPHeaderField: @"Content-Type"];
//create the body
NSMutableData *postBody = [NSMutableData data];
[postBody appendData:[[NSString stringWithFormat:@"<xml>"] dataUsingEncoding:NSUTF8StringEncoding]];
[postBody appendData:[[NSString stringWithFormat:@"<yourcode/>"] dataUsingEncoding:NSUTF8StringEncoding]];
[postBody appendData:[[NSString stringWithFormat:@"</xml>"] dataUsingEncoding:NSUTF8StringEncoding]];
//post
[request setHTTPBody:postBody];
//get response
NSHTTPURLResponse* urlResponse = nil;
NSError *error = [[NSError alloc] init];
NSData *responseData = [NSURLConnection sendSynchronousRequest:request returningResponse:&urlResponse error:&error];
NSString *result = [[NSString alloc] initWithData:responseData encoding:NSUTF8StringEncoding];
NSLog(@"Response Code: %d", [urlResponse statusCode]);
if ([urlResponse statusCode] >= 200 && [urlResponse statusCode] < 300) {
NSLog(@"Response: %@", result);
//here you get the response
}当然，您也可以使用异步请求。然后你必须实现委托。
编辑。您还可以尝试执行以下操作：
NSString* boundary = @"---------------------------14737809831466499882746641449";
NSMutableData* postbody = [NSMutableData dataWithCapacity: 200];
[postbody appendData:[[NSString stringWithFormat:@"\r\n--%@\r\n", _boundary] dataUsingEncoding:NSUTF8StringEncoding]];
[postbody appendData:[[NSString stringWithFormat:@"Content-Disposition: form-data; name=\"%@\"; filename=\"content.xml\"\r\n", name] dataUsingEncoding:NSUTF8StringEncoding]];
[postbody appendData:[[NSString stringWithString:@"Content-Type: text/xml\r\n\r\n"] dataUsingEncoding:NSUTF8StringEncoding]];
[postbody appendData:[[NSString stringWithFormat:@"%@\r\n", val] dataUsingEncoding: NSUTF8StringEncoding]];
NSMutableURLRequest* requestURL= [[[NSMutableURLRequest alloc] init] autorelease];
[requestURL setURL:[NSURL URLWithString: _request.text]];
[requestURL setHTTPMethod:@"POST"];
NSString *contentType = [NSString stringWithFormat:@"multipart/form-data; boundary=%@", boundary];
[requestURL addValue:contentType forHTTPHeaderField: @"Content-Type"];
[requestURL addValue: [NSString stringWithFormat:@"%d",[postbody length]] forHTTPHeaderField: @"Content-length"];
[requestURL setHTTPBody: postbody];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/4918205
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