how to pass arguments to url (.php file)

Question

http://something.php?ownerID=13&view=userAgreement i need to pass two variables to this .. how can i pass these i am declared ownerid to 13 in my config playlist and i am able to get it ,how to set "view = useragreement" string to view variable....

can any body show me sample code.. i am using like

NSString* termsURL = [[[NSBundle mainBundle] infoDictionary] objectForKey:@"termsURL"];
 NSURL *url = [NSURL URLWithString:[NSString stringWithFormat:@"%@%@%@%@%@", termsURL, @"?ownerID=", ownerID,@"?view=",userAgreement]];

but is not working,i am checking parsing is success or not with this url but is failing...

Answer 1

Code taken straight from my app. It works for what you need

NSString *post = [NSString stringWithFormat:@"ownerID=13&view=userAgreement"];

NSData *postData = [post dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES];

NSString *postLength = [NSString stringWithFormat:@"%d", [postData length]];

//set up the request to the website
NSMutableURLRequest *request = [[[NSMutableURLRequest alloc] init] autorelease];

[request setURL:[NSURL URLWithString:@"http://something.php"]];
[request setHTTPMethod:@"POST"];
[request setValue:postLength forHTTPHeaderField:@"Content-Length"];
[request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"];

[request setHTTPBody:postData];
NSError *error;
NSURLResponse *response;

NSData *urlData=[NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error];
NSString *result3=[[[NSString alloc]initWithData:urlData encoding:NSUTF8StringEncoding]autorelease];