Question

在Java中，当HTTP结果为404范围时，此代码抛出异常：

URL url = new URL("http://stackoverflow.com/asdf404notfound");
HttpURLConnection conn = (HttpURLConnection) url.openConnection();
conn.getInputStream(); // throws!

在我的情况下，我碰巧知道内容是404，但我仍然想阅读回复的正文。

（在我的实际案例中，响应代码是403，但是响应正文解释了拒绝的原因，我想将其显示给用户。）

如何访问响应正文？

Answer 1

Here is the bug report（关闭，不会修复，不是错误）。

他们的建议就是编码：

HttpURLConnection httpConn = (HttpURLConnection)_urlConnection;
InputStream _is;
if (httpConn.getResponseCode() < HttpURLConnection.HTTP_BAD_REQUEST) {
    _is = httpConn.getInputStream();
} else {
     /* error from server */
    _is = httpConn.getErrorStream();
}

Answer 2

这与我遇到的问题相同：如果您尝试从连接中读取HttpUrlConnection，则FileNotFoundException会返回getInputStream() 当状态代码高于400时，您应该使用getErrorStream()。

除此之外，请注意，因为成功状态代码不仅仅是200，即使是201,204等也经常被用作成功状态。

以下是我如何管理它的示例

... connection code code code ...

// Get the response code 
int statusCode = connection.getResponseCode();

InputStream is = null;

if (statusCode >= 200 && statusCode < 400) {
   // Create an InputStream in order to extract the response object
   is = connection.getInputStream();
}
else {
   is = connection.getErrorStream();
}

... callback/response to your handler....

通过这种方式，您将能够在成功和错误情况下获得所需的响应。

希望这有帮助！

Answer 3

在.Net中，您拥有WebException的Response属性，该属性允许在异常时访问流。所以我想这对Java来说是个好方法......

private InputStream dispatch(HttpURLConnection http) throws Exception {
    try {
        return http.getInputStream();
    } catch(Exception ex) {
        return http.getErrorStream();
    }
}

或者我使用的实现。（可能需要更改编码或其他内容。在当前环境中工作。）

private String dispatch(HttpURLConnection http) throws Exception {
    try {
        return readStream(http.getInputStream());
    } catch(Exception ex) {
        readAndThrowError(http);
        return null; // <- never gets here, previous statement throws an error
    }
}

private void readAndThrowError(HttpURLConnection http) throws Exception {
    if (http.getContentLengthLong() > 0 && http.getContentType().contains("application/json")) {
        String json = this.readStream(http.getErrorStream());
        Object oson = this.mapper.readValue(json, Object.class);
        json = this.mapper.writer().withDefaultPrettyPrinter().writeValueAsString(oson);
        throw new IllegalStateException(http.getResponseCode() + " " + http.getResponseMessage() + "\n" + json);
    } else {
        throw new IllegalStateException(http.getResponseCode() + " " + http.getResponseMessage());
    }
}

private String readStream(InputStream stream) throws Exception {
    StringBuilder builder = new StringBuilder();
    try (BufferedReader in = new BufferedReader(new InputStreamReader(stream))) {
        String line;
        while ((line = in.readLine()) != null) {
            builder.append(line); // + "\r\n"(no need, json has no line breaks!)
        }
        in.close();
    }
    System.out.println("JSON: " + builder.toString());
    return builder.toString();
}

Answer 4

我知道这不能直接回答这个问题，但是你可能想看看Commons HttpClient，而不是使用Sun提供的HTTP连接库，这在我看来很远更容易使用的API。

Answer 5

首先检查响应代码，然后使用HttpURLConnection.getErrorStream()

Answer 6

InputStream is = null;
if (httpConn.getResponseCode() !=200) {
    is = httpConn.getErrorStream();
} else {
     /* error from server */
    is = httpConn.getInputStream();
}

Answer 7

我的运行代码。

  HttpURLConnection httpConn = (HttpURLConnection) urlConn;    
 if (httpConn.getResponseCode() < HttpURLConnection.HTTP_BAD_REQUEST) {
                        in = new InputStreamReader(urlConn.getInputStream());
                        BufferedReader bufferedReader = new BufferedReader(in);
                        if (bufferedReader != null) {
                            int cp;
                            while ((cp = bufferedReader.read()) != -1) {
                                sb.append((char) cp);
                            }
                            bufferedReader.close();
                        }
                            in.close();

                    } else {
                        /* error from server */
                        in = new InputStreamReader(httpConn.getErrorStream());
                    BufferedReader bufferedReader = new BufferedReader(in);
                    if (bufferedReader != null) {
                        int cp;
                        while ((cp = bufferedReader.read()) != -1) {
                            sb.append((char) cp);
                        }
                        bufferedReader.close();
                    }    
                    in.close();
                    }
                    System.out.println("sb="+sb);

Answer 8

如何在Java中读取404响应正文：

使用Apache库-https://hc.apache.org/httpcomponents-client-4.5.x/httpclient/apidocs/

或 Java 11-https://docs.oracle.com/en/java/javase/11/docs/api/java.net.http/java/net/http/HttpClient.html

下面给出的代码片段使用Apache：

import org.apache.http.impl.client.CloseableHttpClient;
import org.apache.http.impl.client.HttpClients;
import org.apache.http.client.methods.CloseableHttpResponse;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.util.EntityUtils;

CloseableHttpClient client = HttpClients.createDefault();
CloseableHttpResponse resp = client.execute(new HttpGet(domainName + "/blablablabla.html"));
String response = EntityUtils.toString(resp.getEntity());

在Java中读取错误响应主体

8 个答案: