2

我有一个没有使用任何第三方位的工作 http GET，我是 iOS 新手，所以最初设置起来很困难。我的代码如下所示：

-(NSString *) SendGetRequestToRest:(NSString *)urlEndString
{

NSString *userName = @"userN";
NSString *password = @"PassW";

NSString *urlBaseString = @"http://someurl.co.uk/";

NSString *urlString = [NSString stringWithFormat:@"%@%@", urlBaseString, urlEndString];

NSMutableURLRequest *request = [[NSMutableURLRequest alloc] init];

[request setURL:[NSURL URLWithString:urlString]];

[request setHTTPMethod:@"GET"];

NSString *str1 = [NSString stringWithFormat:@"%@:%@", userName, password];
NSString *encodedString = [self stringByBase64EncodingWithString:str1];
[request addValue:[NSString stringWithFormat:@"Basic %@",encodedString] forHTTPHeaderField:@"Authorization"];

NSData *returnData = [NSURLConnection sendSynchronousRequest:request returningResponse:nil error:nil];

NSString *str = [[NSString alloc] initWithData:returnData encoding:NSUTF8StringEncoding];
NSLog(@"str: %@", str);

return str;

}

我需要做的是跟踪 http GET 的状态代码不是很好的 200 时，我看到了如何在 iOS 中检查 Web 服务器的状态？这看起来很有希望，即添加以下内容：

-(void)connection:(NSURLConnection *)connection didReceiveResponse:(NSHTTPURLResponse *)response
{
    if ([response statusCode] == 404)
    {
    /// do some stuff
    }
}

但我看不到如何将其连接到

NSData *returnData = [NSURLConnection sendSynchronousRequest:request returningResponse:nil error:nil];

因为它不接受代表？

4

1 回答 1

12

您可以使用returningResponse参数来获取“响应”：

NSHTTPURLResponse *response = nil;
NSData *returnData = [NSURLConnection sendSynchronousRequest:request returningResponse:&response error:nil];
if ([response statusCode] == 404)
{
   // Do whatever you want to do after getting response
}
于 2013-07-24T09:50:47.393 回答