[iPhone] How to POST url in safari

Question

Hi!
I want my app to open url in Safari and to login.
I know how to just open URL:
UIApplication *app = [UIApplication sharedApplication];
[app openURL:[[NSURL alloc] initWithString:@"http://www.somedomain.net"]];
But there is a login form on that site. Here is it's code:
<form action="http://www.somedomain.net/login.php" method="post">
<td width="68%"><input name="username" class="formlogin" size="23" type="text" /></td>
<td width="10%"> </td>
</tr>
<tr>
<td>Password:</td>
<td><input name="pass" class="formlogin" size="23" type="password" /></td>
it looks like I have to POST params to http://www.somedomain.net/login.php
Well, I know how to POST
NSString *myRequestString = @"username=sampleuser&pass=samplepass";
NSData *myRequestData = [NSData dataWithBytes: [ myRequestString UTF8String ] length: [ myRequestString length ] ];
NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL:[NSURL URLWithString:@"http://somedomain.net/login.php" ]];
[ request setHTTPMethod: @"POST" ];
[ request setHTTPBody: myRequestData ];
NSURLConnection *connection = [[NSURLConnection alloc]
initWithRequest:request
delegate:self];
[connection release];
[request release];
But how to combine it all together? Just POSTing and calling openURL does not work. Is it possible at all? openURL opens safari and GETs? What if a site could log me in on GET request with username and pass as params, would openURL work then?

Answer 1

No, you cannot specify a request method in a URL.