使用HttpRequest.execute()的异常:无效使用SingleClientConnManager:仍然分配了连接

Question

我正在使用google-api-client-java 1.2.1-alpha来执行POST请求,并在执行()HttpRequest时获得以下stacktrace.

它在我捕获并忽略从先前的POST到同一URL的403错误后立即发生,并重新使用传输进行后续请求.(它在一个循环中插入多个条目到同一个ATOM提要).

在403之后,我应该做些什么来"清理"？

Exception in thread "main" java.lang.IllegalStateException: Invalid use of SingleClientConnManager: connection still allocated.
Make sure to release the connection before allocating another one.
    at org.apache.http.impl.conn.SingleClientConnManager.getConnection(SingleClientConnManager.java:199)
    at org.apache.http.impl.conn.SingleClientConnManager$1.getConnection(SingleClientConnManager.java:173)
    at org.apache.http.impl.client.DefaultRequestDirector.execute(DefaultRequestDirector.java:390)
    at org.apache.http.impl.client.AbstractHttpClient.execute(AbstractHttpClient.java:641)
    at org.apache.http.impl.client.AbstractHttpClient.execute(AbstractHttpClient.java:576)
    at org.apache.http.impl.client.AbstractHttpClient.execute(AbstractHttpClient.java:554)
    at com.google.api.client.apache.ApacheHttpRequest.execute(ApacheHttpRequest.java:47)
    at com.google.api.client.http.HttpRequest.execute(HttpRequest.java:207)
    at au.com.machaira.pss.gape.RedirectHandler.execute(RedirectHandler.java:38)
    at au.com.machaira.pss.gape.ss.model.records.TableEntry.executeModification(TableEntry.java:81)

为什么我下面的代码会尝试获取新连接？

Answer 1

在为另一个请求重用连接之前,您需要使用响应主体.您不仅应该读取响应状态,还应该完全读取InputStream最后一个字节的响应,从而忽略读取的字节.

Answer 2

在使用HttpClient和Jetty构建测试框架时,我遇到了类似的问题.我不得不从我的客户端向Servelet创建多个请求,但它在执行时给出了相同的异常.

我在http://foo.jasonhudgins.com/2010/03/http-connections-revisited.html找到了另一种选择.

您还可以使用以下方法来实例化您的客户端.

public static DefaultHttpClient getThreadSafeClient()  {

    DefaultHttpClient client = new DefaultHttpClient();
    ClientConnectionManager mgr = client.getConnectionManager();
    HttpParams params = client.getParams();
    client = new DefaultHttpClient(new ThreadSafeClientConnManager(params, 

            mgr.getSchemeRegistry()), params);
    return client;
}

Answer 3

类似的异常消息(至少Apache Jarkata Commons HTTP Client 4.2)是:

java.lang.IllegalStateException: Invalid use of BasicClientConnManager: connection still allocated. Make sure to release the connection before allocating another one.

当两个或多个线程与单个线程交互时,可能会发生此异常org.apache.http.impl.client.DefaultHttpClient.

你如何制作一个4.2 DefaultHttpClient实例线程安全(线程安全,因为两个或多个线程可以与它交互而不会得到上面的错误消息)？DefaultHttpClient以ClientConnectionManager形式提供连接池org.apache.http.impl.conn.PoolingClientConnectionManager!

/* using
    <dependency>
        <groupId>org.apache.httpcomponents</groupId>
        <artifactId>httpclient</artifactId>
        <version>4.2.2</version>
    </dependency>
*/

import org.apache.http.HttpResponse;
import org.apache.http.HttpStatus;
import org.apache.http.params.HttpConnectionParams;
import org.apache.http.client.HttpClient;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.impl.conn.PoolingClientConnectionManager;
import org.apache.http.impl.conn.SchemeRegistryFactory;
import org.apache.http.params.HttpParams;
import org.apache.http.client.methods.HttpGet;

public class MyComponent {

    private HttpClient client;

    {
        PoolingClientConnectionManager conMan = new PoolingClientConnectionManager( SchemeRegistryFactory.createDefault() );
        conMan.setMaxTotal(200);
        conMan.setDefaultMaxPerRoute(200);

        client = new DefaultHttpClient(conMan);

        //The following parameter configurations are not
        //neccessary for this example, but they show how
        //to further tweak the HttpClient
        HttpParams params = client.getParams();
        HttpConnectionParams.setConnectionTimeout(params, 20000);
        HttpConnectionParams.setSoTimeout(params, 15000);
    }


    //This method can be called concurrently by several threads
    private InputStream getResource(String uri) {
        try {
            HttpGet method = new HttpGet(uri);
            HttpResponse httpResponse = client.execute(method);
            int statusCode = httpResponse.getStatusLine().getStatusCode();
            InputStream is = null;
            if (HttpStatus.SC_OK == statusCode) {
                logger.debug("200 OK Amazon request");
                is = httpResponse.getEntity().getContent();
            } else {
                logger.debug("Something went wrong, statusCode is {}",
                        statusCode);
                 EntityUtils.consume(httpResponse.getEntity());
            }
            return is;
        } catch (Exception e) {
            logger.error("Something went terribly wrong", e);
            throw new RuntimeException(e);
        }
    }
}

Answer 4

这是一个经常被问到的问题.BalusC的回答是正确的.请捕获HttpReponseException,并调用HttpResponseException.回应.ignore().如果需要阅读错误消息,请使用响应.如果您不知道响应内容类型,则为parseAsString(),否则,如果您确实知道内容类型使用响应.parseAs(MyType.class).

从一个简单的代码片断YouTubeSample.java在YouTube上jsonc样本(尽管通常你会想要做的事聪明在实际应用中):

  } catch (HttpResponseException e) {
    System.err.println(e.response.parseAsString());
  }

完全披露:我是google-api-java-client项目的所有者.