Eu criei um formulário de consulta de usuário no ios. Estou usando o php como servidor. Eu construí uma string de consulta como abaixo no IOS

http://www.mydomain.in/androidmail_enquiry.php?name=Vinoth Kumar&phone=04259280244&[email protected]&address=Coimbatore&comments=Sample Enquiry&mobile=xxxxxxxx

Código:

-(void)sendEnquiryDetails
{
    cmttextview.text = @"Sample Enquiry";
    NSString *siteurl = @"http://www.mydomain.in/androidmail_enquiry.php?";
    NSString *name = txtName.text;
    NSString *phone = txtPhone.text;
    NSString *email = txtEmail.text;
    NSString *address = txtAddress.text;
    NSString *comments = cmttextview.text;
    NSString *mobile = txtMobile.text;
    NSString *enquiryurl = [NSString stringWithFormat:@"%@name=%@&phone=%@&email=%@&address=%@&comments=%@&mobile=%@",siteurl,name,phone,email,address,comments,mobile];

    NSLog(enquiryurl);

   NSURLRequest *request = [NSURLRequest requestWithURL:[NSURL URLWithString:enquiryurl]];
    NSURLConnection *conn = [[NSURLConnection alloc] initWithRequest:request  delegate:self];
}
- (void)connection:(NSURLConnection *)connection didReceiveResponse:(NSURLResponse *)response
{
    //NSLog(@"didReceiveResponse");
    [self.responseData setLength:0];
}

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

-(void)connection:(NSURLConnection *)connection didFailWithError:(NSError *)error
{
    NSLog(@"%@",[NSString stringWithFormat :@"didfailwitherror: %@", [error description]]);
}

-(void)connectionDidFinishLoading: (NSURLConnection *)connection{
    NSLog(@"Success Code:%@",self.responseData);
}

Mas quando eu enviar isso usando NSURLRequest recebendo erro como URL inválido (erro de formação de URL) no método didFailWithError.