135

如何连接到需要身份验证的 Java 中的远程 URL。我正在尝试找到一种方法来修改以下代码，以便能够以编程方式提供用户名/密码，因此它不会抛出 401。

URL url = new URL(String.format("http://%s/manager/list", _host + ":8080"));
HttpURLConnection connection = (HttpURLConnection)url.openConnection();
4

12 回答 12

141

您可以为 http 请求设置默认身份验证器，如下所示：

Authenticator.setDefault (new Authenticator() {
    protected PasswordAuthentication getPasswordAuthentication() {
        return new PasswordAuthentication ("username", "password".toCharArray());
    }
});

此外，如果您需要更大的灵活性，您可以查看Apache HttpClient，它将为您提供更多的身份验证选项（以及会话支持等）

于 2009-01-30T18:51:15.743 回答
139

有一个原生且侵入性较小的替代方案，仅适用于您的呼叫。

URL url = new URL(“location address”);
URLConnection uc = url.openConnection();
String userpass = username + ":" + password;
String basicAuth = "Basic " + new String(Base64.getEncoder().encode(userpass.getBytes()));
uc.setRequestProperty ("Authorization", basicAuth);
InputStream in = uc.getInputStream();
于 2011-02-28T01:42:29.887 回答
81

您还可以使用以下内容，不需要使用外部包：

URL url = new URL(“location address”);
URLConnection uc = url.openConnection();

String userpass = username + ":" + password;
String basicAuth = "Basic " + javax.xml.bind.DatatypeConverter.printBase64Binary(userpass.getBytes());

uc.setRequestProperty ("Authorization", basicAuth);
InputStream in = uc.getInputStream();
于 2012-09-26T14:08:25.557 回答
46

如果您在协议和域之间输入用户名和密码时使用正常登录，这更简单。它也适用于登录和不登录。

示例网址： http://user:pass@domain.com/url

URL url = new URL("http://user:pass@domain.com/url");
URLConnection urlConnection = url.openConnection();

if (url.getUserInfo() != null) {
    String basicAuth = "Basic " + new String(new Base64().encode(url.getUserInfo().getBytes()));
    urlConnection.setRequestProperty("Authorization", basicAuth);
}

InputStream inputStream = urlConnection.getInputStream();

请在 valerybodak 的评论中注意它是如何在 Android 开发环境中完成的。

于 2012-10-29T13:00:12.447 回答
8

当我来到这里寻找 Android-Java-Answer 时，我将做一个简短的总结：

  1. 使用James van Huis所示的java.net.Authenticator
  2. 使用Apache Commons HTTP Client，如this Answer
  3. 使用基本的java.net.URLConnection并手动设置 Authentication-Header，如下所示

如果您想在Android中使用java.net.URLConnection和基本身份验证，请尝试以下代码：

URL url = new URL("http://www.mywebsite.com/resource");
URLConnection urlConnection = url.openConnection();
String header = "Basic " + new String(android.util.Base64.encode("user:pass".getBytes(), android.util.Base64.NO_WRAP));
urlConnection.addRequestProperty("Authorization", header);
// go on setting more request headers, reading the response, etc
于 2014-12-06T10:42:41.373 回答
7

能够使用 HttpsURLConnection 设置身份验证

           URL myUrl = new URL(httpsURL);
            HttpsURLConnection conn = (HttpsURLConnection)myUrl.openConnection();
            String userpass = username + ":" + password;
            String basicAuth = "Basic " + new String(Base64.getEncoder().encode(userpass.getBytes()));
            //httpsurlconnection
            conn.setRequestProperty("Authorization", basicAuth);

从这篇文章中获取的更改很少。Base64 来自 java.util 包。

于 2019-03-26T18:53:20.113 回答
4

小心使用“Base64().encode()”方法，我和我的团队遇到了 400 个 Apache 错误请求问题，因为它在生成的字符串末尾添加了一个 \r\n。

由于 Wireshark，我们发现它可以嗅探数据包。

这是我们的解决方案：

import org.apache.commons.codec.binary.Base64;

HttpGet getRequest = new HttpGet(endpoint);
getRequest.addHeader("Authorization", "Basic " + getBasicAuthenticationEncoding());

private String getBasicAuthenticationEncoding() {

        String userPassword = username + ":" + password;
        return new String(Base64.encodeBase64(userPassword.getBytes()));
    }

希望能帮助到你！

于 2014-06-05T22:23:34.403 回答
3

使用此代码进行基本身份验证。

URL url = new URL(path);
String userPass = "username:password";
String basicAuth = "Basic " + Base64.encodeToString(userPass.getBytes(), Base64.DEFAULT);//or
//String basicAuth = "Basic " + new String(Base64.encode(userPass.getBytes(), Base64.No_WRAP));
HttpURLConnection urlConnection = (HttpURLConnection)url.openConnection();
urlConnection.setRequestProperty("Authorization", basicAuth);
urlConnection.connect();

于 2017-05-29T07:58:00.860 回答
3

对于您无法控制打开连接的代码的情况，我想提供一个答案。就像我使用URLClassLoader从受密码保护的服务器加载 jar 文件时所做的那样。

该Authenticator解决方案可以工作，但缺点是它首先尝试在没有密码的情况下访问服务器，并且只有在服务器要求输入密码后才提供密码。如果您已经知道服务器需要密码，那将是不必要的往返。

public class MyStreamHandlerFactory implements URLStreamHandlerFactory {

    private final ServerInfo serverInfo;

    public MyStreamHandlerFactory(ServerInfo serverInfo) {
        this.serverInfo = serverInfo;
    }

    @Override
    public URLStreamHandler createURLStreamHandler(String protocol) {
        switch (protocol) {
            case "my":
                return new MyStreamHandler(serverInfo);
            default:
                return null;
        }
    }

}

public class MyStreamHandler extends URLStreamHandler {

    private final String encodedCredentials;

    public MyStreamHandler(ServerInfo serverInfo) {
        String strCredentials = serverInfo.getUsername() + ":" + serverInfo.getPassword();
        this.encodedCredentials = Base64.getEncoder().encodeToString(strCredentials.getBytes());
    }

    @Override
    protected URLConnection openConnection(URL url) throws IOException {
        String authority = url.getAuthority();
        String protocol = "http";
        URL directUrl = new URL(protocol, url.getHost(), url.getPort(), url.getFile());

        HttpURLConnection connection = (HttpURLConnection) directUrl.openConnection();
        connection.setRequestProperty("Authorization", "Basic " + encodedCredentials);

        return connection;
    }

}

这将注册一个新协议，当添加凭据时该协议my将被替换。http因此，在创建新的时URLClassLoader只需替换http为my，一切都很好。我知道URLClassLoader提供了一个构造函数，URLStreamHandlerFactory但如果 URL 指向 jar 文件，则不使用此工厂。

于 2018-01-01T15:48:33.130 回答
2

从 Java 9 开始，您可以这样做

URL url = new URL("http://www.example.com");
HttpURLConnection connection = (HttpURLConnection)url.openConnection();
connection.setAuthenticator(new Authenticator() {
    protected PasswordAuthentication getPasswordAuthentication() {
        return new PasswordAuthentication ("USER", "PASS".toCharArray());
    }
});
于 2018-07-19T11:22:58.843 回答
1

我是这样做的，您需要这样做，只需复制粘贴就可以了

    HttpURLConnection urlConnection;
    String url;
 //   String data = json;
    String result = null;
    try {
        String username ="danish.hussain@gmail.com";
        String password = "12345678";

        String auth =new String(username + ":" + password);
        byte[] data1 = auth.getBytes(UTF_8);
        String base64 = Base64.encodeToString(data1, Base64.NO_WRAP);
        //Connect
        urlConnection = (HttpURLConnection) ((new URL(urlBasePath).openConnection()));
        urlConnection.setDoOutput(true);
        urlConnection.setRequestProperty("Content-Type", "application/json");
        urlConnection.setRequestProperty("Authorization", "Basic "+base64);
        urlConnection.setRequestProperty("Accept", "application/json");
        urlConnection.setRequestMethod("POST");
        urlConnection.setConnectTimeout(10000);
        urlConnection.connect();
        JSONObject obj = new JSONObject();

        obj.put("MobileNumber", "+97333746934");
        obj.put("EmailAddress", "danish.hussain@mee.com");
        obj.put("FirstName", "Danish");
        obj.put("LastName", "Hussain");
        obj.put("Country", "BH");
        obj.put("Language", "EN");
        String data = obj.toString();
        //Write
        OutputStream outputStream = urlConnection.getOutputStream();
        BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(outputStream, "UTF-8"));
        writer.write(data);
        writer.close();
        outputStream.close();
        int responseCode=urlConnection.getResponseCode();
        if (responseCode == HttpsURLConnection.HTTP_OK) {
            //Read
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(urlConnection.getInputStream(), "UTF-8"));

        String line = null;
        StringBuilder sb = new StringBuilder();

        while ((line = bufferedReader.readLine()) != null) {
            sb.append(line);
        }

        bufferedReader.close();
        result = sb.toString();

        }else {
        //    return new String("false : "+responseCode);
        new String("false : "+responseCode);
        }

    } catch (UnsupportedEncodingException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    } catch (JSONException e) {
        e.printStackTrace();
    }
于 2018-04-05T07:57:23.290 回答
0

ANDROD 实现 一种从 Web 服务请求数据/字符串响应的完整方法，请求使用用户名和密码进行授权

public static String getData(String uri, String userName, String userPassword) {
        BufferedReader reader = null;
        byte[] loginBytes = (userName + ":" + userPassword).getBytes();

        StringBuilder loginBuilder = new StringBuilder()
                .append("Basic ")
                .append(Base64.encodeToString(loginBytes, Base64.DEFAULT));

        try {
            URL url = new URL(uri);
            HttpURLConnection connection = (HttpURLConnection) url.openConnection();
            connection.addRequestProperty("Authorization", loginBuilder.toString());

            StringBuilder sb = new StringBuilder();
            reader = new BufferedReader(new InputStreamReader(connection.getInputStream()));
            String line;
            while ((line = reader.readLine())!= null){
                sb.append(line);
                sb.append("\n");
            }

            return  sb.toString();

        } catch (Exception e) {
            e.printStackTrace();
            return null;
        } finally {
            if (null != reader){
                try {
                    reader.close();
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
        }
    }
于 2016-12-15T23:04:18.030 回答