如何使用 php 网络服务成功登录并移动到另一个视图

Question

首先我有两个文本字段，第一个是登录名，第二个是密码和一个登录按钮。我正在使用通过推送连接连接到另一个视图控制器的故事板和登录按钮。这次在我的项目中工作，在移动另一个视图后将用户名和密码放在文本字段和 select 登录按钮中。我输入了错误的密码登录仍然和移动另一个视图。我认为某处条件不对。请帮助我到底该怎么做。谢谢

我按照这个教程Dipin Krishna Tutorial

我的密码是

- (IBAction)loginac:(id)sender {
    NSInteger success = 0;
    @try {
        
        if([[self.txfld text] isEqualToString:@""] || [[self.tx1 text] isEqualToString:@""] ) {
            
            [self alertStatus:@"Please enter Email and Password" :@"Sign in Failed!" :0];
            
        } else {
            NSString *post =[[NSString alloc] initWithFormat:@"username=%@&password=%@",[self.txfld text],[self.tx1 text]];
            NSLog(@"PostData: %@",post);
            
            NSURL *url=[NSURL URLWithString:@"https://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];
}

Answer 1

删除此

if (success) {
[self performSegueWithIdentifier:@"login_success" sender:self];

}

并添加到 SUCCESS 条件内部，肯定有效

 if(success == 1)
        {
            NSLog(@"Login SUCCESS");
           [self performSegueWithIdentifier:@"login_success" sender:self];
        }

完整答案

- (IBAction)loginac:(id)sender {


    if([[self.txfld text] isEqualToString:@""] || [[self.tx1 text] isEqualToString:@""] ) {

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

    } else {
        NSString *post =[[NSString alloc] initWithFormat:@"username=%@&password=%@",[self.txfld text],[self.tx1 text]];
        NSLog(@"PostData: %@",post);

        NSURL *url=[NSURL URLWithString:@"https://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)
        {


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

          int success = [jsonData[@"success"] integerValue];


            if(success == 1)
            {
                NSLog(@"Login SUCCESS");
                [self performSegueWithIdentifier:@"login_success" sender:self];
            } 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];
        }
    }



}

如何使用 php 网络服务成功登录并移动到另一个视图

How to successfully login and move to another view using php web service

json

objective-c

nsurlrequest

ios

nsurlsession

我的密码是