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我想在 URL（Web 服务）中传递用户名和密码以进行用户身份验证，这将返回 true 和 false。我这样做如下：

NSString *userName = [NSString stringWithFormat:@"parameterUser=%@",txtUserName];
NSString *passWord = [NSString stringWithFormat:@"parameterPass=%@",txtPassword];
NSData *getUserData = [userName dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES];
NSString *getUserLength = [NSString stringWithFormat:@"%d",[getUserData length]];
NSData *getPassData = [passWord dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES];
NSString *getPassLength = [NSString stringWithFormat:@"%d",[getPassData length]];
NSMutableURLRequest *request = [[NSMutableURLRequest alloc]init];
[request setURL:[NSURL URLWithString:@"http://URL/service1.asmx"]];
[request setHTTPMethod:@"GET"];

现在，我想知道如何在这个 URL 中传递我的用户名和密码来发出请求。有人可以建议或提供一些示例代码吗？谢谢。

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4 回答 4

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尝试这个 ：-

NSString *userName = [NSString stringWithFormat:@"parameterUser=%@",txtUserName.text];
    NSString *passWord = [NSString stringWithFormat:@"parameterPass=%@",txtPassword.text];
    NSData *getUserData = [userName dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES];
    NSString *getUserLength = [NSString stringWithFormat:@"%d",[getUserData length]];
    NSData *getPassData = [passWord dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES];
    NSString *getPassLength = [NSString stringWithFormat:@"%d",[getPassData length]];
    NSMutableURLRequest *request = [[NSMutableURLRequest alloc]init];
    [request setURL:[NSURL URLWithString:[NSString stringWithFormat:@"http://URL/service1.asmx?%@&%@",userName,passWord]]];
    [request setHTTPMethod:@"GET"];

希望对你有帮助。。

于 2013-02-11T13:26:19.397 回答
0 
NSString *urlStr = [NSString stringWithFormat:@"http://URL/service1.asmx?%@&%@",userName,passWord];
[request setURL:[NSURL URLWithString:urlStr]];
于 2013-02-11T13:28:00.693 回答
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首先，我不会在 url 中传递用户名和密码。您应该使用 post 来执行此操作。 

NSURL *url = [NSURL URLWithString:[NSString stringWithFormat:@"http://URL/service1.asmx?"]];

NSString *userName = [NSString stringWithFormat:@"parameterUser=%@",txtUserName];
NSString *passWord = [NSString stringWithFormat:@"parameterPass=%@",txtPassword];
NSString *postString = [NSString stringWithFormat:@"username=%@&password=%@",userName, passWord];
NSData *postData = [NSData dataWithBytes: [postString UTF8String] length: [postString length]];

//URL Requst Object
NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:url cachePolicy:NSURLRequestReloadIgnoringCacheData timeoutInterval:TIMEOUT];
[request setHTTPMethod:@"POST"];
[request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"];
[request setHTTPBody: postData];

这比在 url 中传递敏感数据更安全。

编辑

要获得回复，您可以查看此内容。NSURLConnection和AppleDoc NSURLConnection

您可以使用几种不同的方法来处理来自服务器的响应。您可以使用NSURLConnectionDelegate

NSURLConnection *connection = [[NSURLConnection alloc] initWithRequest:request delegate:self];
[self.connection start];

连同委托回调：

- (void)connection:(NSURLConnection *)connection didReceiveData:(NSData *)data {

    NSLog(@"didReceiveData");
    if (!self.receivedData){
        self.receivedData = [NSMutableData data];
    }
    [self.receivedData appendData:data];
}

- (void)connectionDidFinishLoading:(NSURLConnection *)connection {

    NSLog(@"connectionDidFinishLoading");
    NSString *receivedString = [[NSString alloc] initWithData:data encoding:NSUTF8StringEncoding];
    NSLog(@"receivedString:%@",receivedString);

}

或者你也可以使用NSURLConnection sendAsynchronousRequest块

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

[NSURLConnection sendAsynchronousRequest:urlRequest queue:queue completionHandler:^(NSURLResponse *response, NSData *data, NSError *error)
{
    NSString *receivedString = [[NSString alloc] initWithData:data encoding:NSUTF8StringEncoding];
        NSLog(@"receivedString:%@",receivedString);

}];
于 2013-02-11T13:53:17.127 回答
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为了提高安全性，您可以使用 Http Basic Authentication。这里有答案。

于 2013-02-11T14:32:02.840 回答