Question

嗨，我对Xcode很新，我试图使用这个教程登录一个网站的网站是ASPX它有一个简单的用户名和密码字段，但无论我做什么我都无法登录

任何建议都会非常感谢。

基本上我想要这段代码：

(IBAction)sigininClicked:(id)sender {
NSInteger success = 0;
@try {

    if([[self.txtUsername text] isEqualToString:@""] || [[self.txtPassword text] isEqualToString:@""] ) {

        [self alertStatus:@"Please enter Email and Password" :@"Sign in Failed!" :0];

    } else {
        NSString *post =[[NSString alloc] initWithFormat:@"username=%@&password=%@",[self.txtUsername text],[self.txtPassword text]];
        NSLog(@"PostData: %@",post);

        NSURL *url=[NSURL URLWithString:@"http://dipinkrishna.com/jsonlogin.php"];

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

        NSString *postLength = [NSString stringWithFormat:@"%lu", (unsigned long)[postData length]];

        NSMutableURLRequest *request = [[NSMutableURLRequest alloc] init];
        [request setURL:url];
        [request setHTTPMethod:@"POST"];
        [request setValue:postLength forHTTPHeaderField:@"Content-Length"];
        [request setValue:@"application/json" forHTTPHeaderField:@"Accept"];
        [request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"];
        [request setHTTPBody:postData];

        //[NSURLRequest setAllowsAnyHTTPSCertificate:YES forHost:[url host]];

        NSError *error = [[NSError alloc] init];
        NSHTTPURLResponse *response = nil;
        NSData *urlData=[NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error];

        NSLog(@"Response code: %ld", (long)[response statusCode]);

        if ([response statusCode] >= 200 && [response statusCode] < 300)
        {
            NSString *responseData = [[NSString alloc]initWithData:urlData encoding:NSUTF8StringEncoding];
            NSLog(@"Response ==> %@", responseData);

            NSError *error = nil;
            NSDictionary *jsonData = [NSJSONSerialization
                                      JSONObjectWithData:urlData
                                      options:NSJSONReadingMutableContainers
                                      error:&error];

            success = [jsonData[@"success"] integerValue];
            NSLog(@"Success: %ld",(long)success);

            if(success == 1)
            {
                NSLog(@"Login SUCCESS");
            } else {

                NSString *error_msg = (NSString *) jsonData[@"error_message"];
                [self alertStatus:error_msg :@"Sign in Failed!" :0];
            }

        } else {
            //if (error) NSLog(@"Error: %@", error);
            [self alertStatus:@"Connection Failed" :@"Sign in Failed!" :0];
        }
    }
}
@catch (NSException * e) {
    NSLog(@"Exception: %@", e);
    [self alertStatus:@"Sign in Failed." :@"Error!" :0];
}
if (success) {
    [self performSegueWithIdentifier:@"login_success" sender:self];
}
}

- (void) alertStatus:(NSString *)msg :(NSString *)title :(int) tag
{
    UIAlertView *alertView = [[UIAlertView alloc] initWithTitle:title
                                                    message:msg
                                                   delegate:self
                                          cancelButtonTitle:@"Ok"
                                          otherButtonTitles:nil, nil];
alertView.tag = tag;
[alertView show];
}

- (IBAction)backgroundTap:(id)sender {
[self.view endEditing:YES];
}

-(BOOL) textFieldShouldReturn:(UITextField *)textField {
[textField resignFirstResponder];
return YES;
}

@end

要登录此网站LOGIN LINK

Answer 1

首先，您需要在登录页面（网页）中检查请求。看看它，看看那里的登录逻辑。检查这些：它是否也使用json，它是否使用＆＃34;接受＆＃34;作为json数据的HTTPHeaderField？是否将用户名和密码放在httpbody中作为这种格式 - ＆gt; ＆＃34;用户名=％@＆安培;密码=％@＆＃34 ;?发送请求时密码是否应加密？你应该使用哪个网址？

通常，您应该在代码中更改请求部分以满足登录的特定API。

网站登录，ASPX登录页面从应用程序

1 个答案: