java - 为 HttpURLConnection 添加标头

Question

我正在尝试为我的请求添加标头，HttpUrlConnection但该方法setRequestProperty()似乎不起作用。服务器端没有收到任何带有我的标头的请求。

HttpURLConnection hc;
    try {
        String authorization = "";
        URL address = new URL(url);
        hc = (HttpURLConnection) address.openConnection();


        hc.setDoOutput(true);
        hc.setDoInput(true);
        hc.setUseCaches(false);

        if (username != null && password != null) {
            authorization = username + ":" + password;
        }

        if (authorization != null) {
            byte[] encodedBytes;
            encodedBytes = Base64.encode(authorization.getBytes(), 0);
            authorization = "Basic " + encodedBytes;
            hc.setRequestProperty("Authorization", authorization);
        }

score 481 · Accepted Answer

我过去使用过以下代码，它与在 TomCat 中启用的基本身份验证一起工作：

URL myURL = new URL(serviceURL);
HttpURLConnection myURLConnection = (HttpURLConnection)myURL.openConnection();

String userCredentials = "username:password";
String basicAuth = "Basic " + new String(Base64.getEncoder().encode(userCredentials.getBytes()));

myURLConnection.setRequestProperty ("Authorization", basicAuth);
myURLConnection.setRequestMethod("POST");
myURLConnection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
myURLConnection.setRequestProperty("Content-Length", "" + postData.getBytes().length);
myURLConnection.setRequestProperty("Content-Language", "en-US");
myURLConnection.setUseCaches(false);
myURLConnection.setDoInput(true);
myURLConnection.setDoOutput(true);

你可以试试上面的代码。上面的代码是 POST 的，你可以修改它为 GET

score 18 · Accepted Answer

只是因为我在上面的答案中没有看到这一点信息，最初发布的代码片段无法正常工作的原因是因为encodedBytes变量是 abyte[]而不是String值。如果您将 a 传递byte[]给new String()如下，则代码片段可以完美运行。

encodedBytes = Base64.encode(authorization.getBytes(), 0);
authorization = "Basic " + new String(encodedBytes);

score 13 · Accepted Answer

如果您使用的是 Java 8，请使用以下代码。

URLConnection connection = url.openConnection();
HttpURLConnection httpConn = (HttpURLConnection) connection;

String basicAuth = Base64.getEncoder().encodeToString((username+":"+password).getBytes(StandardCharsets.UTF_8));
httpConn.setRequestProperty ("Authorization", "Basic "+basicAuth);

score 6 · Accepted Answer

最后这对我有用

private String buildBasicAuthorizationString(String username, String password) {

    String credentials = username + ":" + password;
    return "Basic " + new String(Base64.encode(credentials.getBytes(), Base64.NO_WRAP));
}

score 5 · Accepted Answer

您的代码很好。您也可以通过这种方式使用相同的东西。

public static String getResponseFromJsonURL(String url) {
    String jsonResponse = null;
    if (CommonUtility.isNotEmpty(url)) {
        try {
            /************** For getting response from HTTP URL start ***************/
            URL object = new URL(url);

            HttpURLConnection connection = (HttpURLConnection) object
                    .openConnection();
            // int timeOut = connection.getReadTimeout();
            connection.setReadTimeout(60 * 1000);
            connection.setConnectTimeout(60 * 1000);
            String authorization="xyz:xyz$123";
            String encodedAuth="Basic "+Base64.encode(authorization.getBytes());
            connection.setRequestProperty("Authorization", encodedAuth);
            int responseCode = connection.getResponseCode();
            //String responseMsg = connection.getResponseMessage();

            if (responseCode == 200) {
                InputStream inputStr = connection.getInputStream();
                String encoding = connection.getContentEncoding() == null ? "UTF-8"
                        : connection.getContentEncoding();
                jsonResponse = IOUtils.toString(inputStr, encoding);
                /************** For getting response from HTTP URL end ***************/

            }
        } catch (Exception e) {
            e.printStackTrace();

        }
    }
    return jsonResponse;
}

如果授权成功，则返回响应码 200

score 1 · Accepted Answer

使用RestAssurd，您还可以执行以下操作：

String path = baseApiUrl; //This is the base url of the API tested
    URL url = new URL(path);
    given(). //Rest Assured syntax 
            contentType("application/json"). //API content type
            given().header("headerName", "headerValue"). //Some API contains headers to run with the API 
            when().
            get(url).
            then().
            statusCode(200); //Assert that the response is 200 - OK

score -1 · Accepted Answer

第一步：获取 HttpURLConnection 对象

URL url = new URL(urlToConnect);
HttpURLConnection httpUrlConnection = (HttpURLConnection) url.openConnection();

第 2 步：使用 setRequestProperty 方法将标头添加到 HttpURLConnection。

Map<String, String> headers = new HashMap<>();

headers.put("X-CSRF-Token", "fetch");
headers.put("content-type", "application/json");

for (String headerKey : headers.keySet()) {
    httpUrlConnection.setRequestProperty(headerKey, headers.get(headerKey));
}

参考链接

java - 为 HttpURLConnection 添加标头

7 回答 7

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