当我将 java 代码传输到有关连接到服务器并获得结果的 iOS 代码时有什么问题？

Question

java代码如下，它即将连接到服务器并获取一些信息，您可以运行java代码并获得｛"carrier"：｛"CFeeC"：1，"COrderC"：2，"CTaskC"：3，"flag"：1｝，"driver"：{"DTaskC":1，"DTraceR"：1、"flag"：1｝，"userinfo"：｛"czBizOrgId"："30500002049，"czWebOrgId"：545587，"hzBizOrgId:"30500000857"，"hzWebOrgId"：545593，"用户电子邮件"："[e-mail]"，"userId"：545566，"userName"："all-all-all"｝，"return_code"："00"，"customer"：｛"flag"：1，"hFeeC"：1、"hOrderA"：1和"hOrderC"：1以及"hOrderS"：1"hPlanM"：1并且"hPlanS"：1}｝｝空

但当我运行iOS代码时，破解了java代码，答案是{"exception"："Authenticated access required"}似乎我发送的参数，服务器没有得到？为什么？

    public static void main(String[] args) {
    System.out.println(test());
}
private static String test() {
    String username = "[e-mail]";
    String password = "1";
        String path = new String("http://61.174.31.28:9091/api/jsonws/UnicomWL-portlet.tcdcolligate/user-login/password/1/email/all%401.com"); // ƴ�����URL��ַ
        //path = path.replace("+","%20");
    try {
        String authString =username + ":" +password;
        String authStringEnc = new String(Base64.encode(authString.getBytes()));
        URL url = new URL(path);
        HttpURLConnection connection = (HttpURLConnection) url.openConnection();
        connection.setRequestMethod("POST");
        connection.setDoOutput(true);
        connection.setRequestProperty("Authorization", "Basic " + authStringEnc);
        BufferedOutputStream bufferedOutputStream = new BufferedOutputStream(
                connection.getOutputStream());
        bufferedOutputStream.flush();
        bufferedOutputStream.close();
        if (connection.getResponseCode() != 200) {
            throw new RuntimeException("ResponseCode != 200");
        }
        BufferedInputStream bufferedInputStream = new BufferedInputStream(
                connection.getInputStream());
        int bytesRead = 0;
        byte[] buffer = new byte[1024];
        while ((bytesRead = bufferedInputStream.read(buffer)) != -1) {

            String chunk = new String(buffer, 0, bytesRead);
            System.out.print(chunk);
        }
        connection.disconnect();
    }
    catch (Exception e) {
    }

    return  null;

但当我尝试使用iOS时，它错了，怎么了？谢谢

    int main(int argc, const char * argv[]) {
@autoreleasepool {
    // insert code here...
    NSLog(@"Hello, World!");
    NSData *authData=[@"[e-mail]:1" dataUsingEncoding:NSUTF8StringEncoding];
    NSString *authStringEnc=[authData base64EncodedStringWithOptions:0];
    NSString *auth=[NSString stringWithFormat:@"Basic +%@",authStringEnc];
    NSLog(@"%@n",auth);
    NSString *path=@"http://61.174.31.28:9091/api/jsonws/UnicomWL-portlet.tcdcolligate/user-login/password/1/email/all%401.com";
    NSURL *url=[NSURL URLWithString:path];
    NSMutableURLRequest *request=[[NSMutableURLRequest alloc] initWithURL:url];
    NSData *returnData=[NSURLConnection sendSynchronousRequest:request returningResponse:nil error:nil];
    NSString *s=[[NSString alloc]initWithData:returnData encoding:NSUTF8StringEncoding];
    NSLog(@"%@",s);
}
return 0;

}

Answer 1

您需要这样做：

NSString *authStr = [NSString stringWithFormat:@"%@:%@", @"username", @"password"];
NSData *authData = [authStr dataUsingEncoding:NSUTF8StringEncoding];
NSString *authValue = [NSString stringWithFormat:@"Basic %@", [authData base64Encoding]];
[request setValue:authValue forHTTPHeaderField:@"Authorization"];

这是一个行之有效的示例项目。样本