在URL进行身份验证通过用户名和密码(Pass username and password in U

我想OT传递URL（Web服务）的用户名和密码进行用户身份验证，这将返回true和false.I'm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，我想知道我怎样才能把我的用户名和密码在这个网址进行申请。 可以在任何一个请建议，或给一些示例代码？ 谢谢。

Answer 1:

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"];

希望它可以帮助你..

Answer 2:

NSString *urlStr = [NSString stringWithFormat:@"http://URL/service1.asmx?%@&%@",userName,passWord];
[request setURL:[NSURL URLWithString:urlStr]];

Answer 3:

为了提高安全性，您可以使用HTTP 基本身份验证 。 有回答这里 。

Answer 4:

首先，我不会在URL跨越传递一个用户名和密码。 你应该这样做使用后。

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];

这是更安全的，然后传递敏感数据跨越在URL中。

编辑

获得响应，你可以检查了这一点。 NSURLConnection的和AppleDoc NSURLConnection的

你可以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];

随着委托回调：

- (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);

}];

文章来源: Pass username and password in URL for authentication