将此 .php 转换为 ObjC 可运行的 NSURLSession POST 请求？

Question

基本上，我正在尝试登录这个website called Mistar。我可以用 php 做到这一点：

<form action="https://mistar.oakland.k12.mi.us/novi/StudentPortal/Home/Login" method="post">
<input type=text name="Pin">
<input type=password name="Password">
<input type="submit" id="LoginButton">
</form>

所以 php（当你运行它时）可以工作。您使用 Pin (20005012) 和密码 (wildcats) 进行身份验证并返回带有 {1, User Authenticated} 之类的页面

现在我要做的是从 iPhone 应用程序（所以 ObjC）登录网站，可能使用 NSURLSession 或其他东西。

这是我目前所拥有的，但它一直给我一个登录页面错误：

- (void)loginToMistar {

    //Create POST request
    NSMutableURLRequest *request = [[NSMutableURLRequest alloc] init];

    //Create and send request
    NSURL *url = [NSURL URLWithString:@"https://mistar.oakland.k12.mi.us/novi/StudentPortal/Home/Login"];
    NSURLRequest *urlRequest = [NSURLRequest requestWithURL:url];
    NSString *postString = [NSString stringWithFormat:@"<input type=text name='Pin'> <input type=password name='Password'> <input type='submit' id='LoginButton'>"];
    NSData * postBody = [postString dataUsingEncoding:NSUTF8StringEncoding];
    [request setHTTPBody:postBody];

 //   [request addValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"];

    NSOperationQueue *queue = [[NSOperationQueue alloc] init];

    NSString* url=[NSString stringWithFormat:url];

    [NSURLConnection sendAsynchronousRequest:urlRequest queue:queue completionHandler:^(NSURLResponse *response, NSData *data, NSError *error)
    {
        // do whatever with the data...and errors
        if ([data length] > 0 && error == nil) {
            NSString *loggedInPage = [[NSString alloc] initWithData:data encoding:NSUTF8StringEncoding];
            NSLog(loggedInPage);
        }
        else {
            NSLog(@"error");
        }
    }];

谁能告诉我代码中的问题出在哪里？

Answer 1

1. 通常，您需要指定请求的Content-type (application/x-www-form-urlencoded)。

2. 您希望构建请求的主体以符合该类型的请求（如 cmorrissey 所说），例如Pin=xxx&Password=yyy。

1234563 /p>

3. 您需要指定您的请求是POST 请求。所以，使用原来的NSMutableURLRequest 并正确配置它，然后停用你的NSURLRequest。

4. 你不希望那一行写着：

   NSString* url=[NSString stringWithFormat:url];
   

5. 您不必创建操作队列。你可以，但是（a）你没有做一些非常慢和计算成本高的事情； (b) 这个完成块很可能最终想要更新 UI，而您永远不会在后台队列上这样做，而只会在主队列上这样做。

6. 由于响应似乎是 JSON，让我们继续解析该响应（如果可以的话）。

因此，您最终会得到以下结果：

- (void)loginToMistarWithPin:(NSString *)pin password:(NSString *)password {

    NSURL *url = [NSURL URLWithString:@"https://mistar.oakland.k12.mi.us/novi/StudentPortal/Home/Login"];

    //Create and send request
    NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL:url];

    [request setHTTPMethod:@"POST"];
    [request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-type"];

    NSString *postString = [NSString stringWithFormat:@"Pin=%@&Password=%@",
                            [self percentEscapeString:pin],
                            [self percentEscapeString:password]];
    NSData * postBody = [postString dataUsingEncoding:NSUTF8StringEncoding];
    [request setHTTPBody:postBody];

    [NSURLConnection sendAsynchronousRequest:request queue:[NSOperationQueue mainQueue] completionHandler:^(NSURLResponse *response, NSData *data, NSError *error)
     {
         // do whatever with the data...and errors
         if ([data length] > 0 && error == nil) {
             NSError *parseError;
             NSDictionary *responseJSON = [NSJSONSerialization JSONObjectWithData:data options:0 error:&parseError];
             if (responseJSON) {
                 // the response was JSON and we successfully decoded it

                 NSLog(@"Response was = %@", responseJSON);
             } else {
                 // the response was not JSON, so let's see what it was so we can diagnose the issue

                 NSString *loggedInPage = [[NSString alloc] initWithData:data encoding:NSUTF8StringEncoding];
                 NSLog(@"Response was not JSON, it was = %@", loggedInPage);
             }
         }
         else {
             NSLog(@"error: %@", error);
         }
     }];
}

- (NSString *)percentEscapeString:(NSString *)string
{
    NSString *result = CFBridgingRelease(CFURLCreateStringByAddingPercentEscapes(kCFAllocatorDefault,
                                                                                 (CFStringRef)string,
                                                                                 (CFStringRef)@" ",
                                                                                 (CFStringRef)@":/?@!$&'()*+,;=",
                                                                                 kCFStringEncodingUTF8));
    return [result stringByReplacingOccurrencesOfString:@" " withString:@"+"];
}

你可以这样称呼它：

[self loginToMistarWithPin:@"20005012" password:@"wildcats"];

Answer 2

以下行不正确。

NSString *postString = [NSString stringWithFormat:@"<input type=text name='Pin'> <input type=password name='Password'> <input type='submit' id='LoginButton'>"];

"<input type=text name='Pin'> <input type=password name='Password'> <input type='submit' id='LoginButton'>" 不是正确的 POST 字符串

你需要用类似"Pin=20005012&Password=wildcats"的东西替换它