12

通常，我喜欢使用完成处理程序块使用 NSURL 的 sendAsynchronousRequest 类方法“触发并忘记”，但是当需要身份验证时，这似乎不是一个选项。

当使用像这样的完成处理程序样式请求时：

[NSURLConnection sendAsynchronousRequest:[NSURLRequest requestWithURL:[NSURL URLWithString:@"http://www.mysite.com/"]]
                                       queue:[NSOperationQueue mainQueue]
                           completionHandler:^(NSURLResponse *response, NSData *data, NSError *error) {


                            //Do Stuff   

                           }];

处理身份验证的正确方法是什么？我是否需要分配和初始化 NSURLConnection 并设置委托而不是使用此类方法样式？我想我了解如何使用委托函数正确进行身份验证，但我试图弄清楚是否可以将其包含在 completionHandler 块中，或者是否有更好的方法来做到这一点。

- (void)connection:(NSURLConnection *)connection willSendRequestForAuthenticationChallenge:(NSURLAuthenticationChallenge *)challenge
{

    if ([challenge previousFailureCount] > 0) {
        NSLog(@"Authentication Failure");
        [connection cancel];
    }
    else
    {

        NSURLCredential *credential = [NSURLCredential credentialWithUser:self.username
                                                                 password:self.password
                                                              persistence:NSURLCredentialPersistenceForSession];
        [[challenge sender] useCredential:credential forAuthenticationChallenge:challenge];
    }

}
4

3 回答 3

22

我认为 completionHandler 方法是针对基本请求的。也许您可以考虑使用AFNetworking，因为我将它与块方法和身份验证一起使用。

编辑....您是否尝试将身份验证标头添加到 NSURLRequest？创建一个 NSMutableURLRequest：

NSMutableURLRequest *urlRequest = [[NSMutableURLRequest alloc] initWithURL:[NSURL URLWithString:@"http://www.example.com/"]];

并像这样添加身份验证标头：

NSString *basicAuthCredentials = [NSString stringWithFormat:@"%@:%@", userName, password];
NSString *authValue = [NSString stringWithFormat:@"Basic %@", AFBase64EncodedStringFromString(basicAuthCredentials)];
[urlRequest setValue:authValue forHTTPHeaderField:@"Authorization"];

AFBase64EncodedStringFromString 函数是这样的：

static NSString * AFBase64EncodedStringFromString(NSString *string) {
    NSData *data = [NSData dataWithBytes:[string UTF8String] length:[string lengthOfBytesUsingEncoding:NSUTF8StringEncoding]];
    NSUInteger length = [data length];
    NSMutableData *mutableData = [NSMutableData dataWithLength:((length + 2) / 3) * 4];

    uint8_t *input = (uint8_t *)[data bytes];
    uint8_t *output = (uint8_t *)[mutableData mutableBytes];

    for (NSUInteger i = 0; i < length; i += 3) {
        NSUInteger value = 0;
        for (NSUInteger j = i; j < (i + 3); j++) {
            value <<= 8;
            if (j < length) {
                value |= (0xFF & input[j]);
            }
        }

        static uint8_t const kAFBase64EncodingTable[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";

        NSUInteger idx = (i / 3) * 4;
        output[idx + 0] = kAFBase64EncodingTable[(value >> 18) & 0x3F];
        output[idx + 1] = kAFBase64EncodingTable[(value >> 12) & 0x3F];
        output[idx + 2] = (i + 1) < length ? kAFBase64EncodingTable[(value >> 6)  & 0x3F] : '=';
        output[idx + 3] = (i + 2) < length ? kAFBase64EncodingTable[(value >> 0)  & 0x3F] : '=';
    }

    return [[NSString alloc] initWithData:mutableData encoding:NSASCIIStringEncoding];
}

然后调用你之前调用的函数，但使用你的新 NSURLRequest：

[NSURLConnection sendAsynchronousRequest:urlRequest
  queue:[NSOperationQueue mainQueue]
  completionHandler:^(NSURLResponse *response, NSData *data, NSError *error) {

}];
于 2012-10-16T19:24:48.173 回答
21

即使问题已经解决了一段时间，我想添加一个不需要 3rd 方库的解决方案：

//HTTP Authentication
NSString *authStr = [NSString stringWithFormat:@"%@:%@", @"Username", @"Password"]];
NSData *authData = [authStr dataUsingEncoding:NSASCIIStringEncoding];
NSString *authValue = [authData base64Encoding];

//Set up Request:
NSMutableURLRequest *request = [[NSMutableURLRequest alloc]init];
[request setURL:[NSURL URLWithString:url]];
//...
// Pack in the user credentials
[request setValue:[NSString stringWithFormat:@"Basic %@",authValue] forHTTPHeaderField:@"Authorization"];

// Send the asynchronous request as usual
[NSURLConnection sendAsynchronousRequest:request queue:[NSOperationQueue mainQueue] completionHandler:^(NSURLResponse *responseCode, NSData *responseData, NSError *error) {

    // Do stuff..

}
于 2013-11-13T11:28:51.060 回答
1 
NSURLConnection is deprecated in iOS 9, try this code which is from LazyTableImages example from here [https://developer.apple.com/library/ios/samplecode/LazyTableImages/Introduction/Intro.html][1]

    @property (nonatomic, strong) NSURLSessionDataTask *sessionTask;



    //somwere in your code

NSMutableURLRequest *request = [[NSMutableURLRequest alloc]
                                    initWithURL:[NSURL URLWithString:self.postLink] ];

    [request setCachePolicy:NSURLRequestReloadIgnoringLocalCacheData];
    [request setHTTPShouldHandleCookies:NO];
    [request setTimeoutInterval:60];
    [request setHTTPMethod:@"POST"];
    [request addValue:[[NSUserDefaults standardUserDefaults] objectForKey:@"token"]
   forHTTPHeaderField:@"ACCESSTOKEN"];


    self.sessionTask = [[NSURLSession sharedSession] dataTaskWithRequest:request
                                                           completionHandler:^(NSData *data, NSURLResponse *response, NSError *error) {
    //do completion staff here
    }];
于 2015-12-28T15:45:42.323 回答