java – 在HttpClient 4.0中设置HttpParams对象

我正在创建新的HttpClient,通过在它的构造函数中传递ThreadSafeClientConnManager和HttpParams,但它总是抛出错误的请求错误.
为了弄清楚出了什么问题,我调试了它但都是徒劳的.为我提供一些解决方案
这是我的代码块

sc.register(new Scheme("http",PlainSocketFactory.getSocketFactory(),80));
sc.register(new Scheme("https",SSLSocketFactory.getSocketFactory(),443));

HttpParams basicParams = new BasicHttpParams();
ThreadSafeClientConnManager connmgr = new ThreadSafeClientConnManager(basicParams,sc);
ConnManagerParams.setMaxConnectionsPerRoute(
    basicParams,// if we have more than 5 concurrent leads,good problem to have
    new ConnPerRoute() {
        public int getMaxForRoute(HttpRoute httproute) {
        return 5;
    }
}); 

g_client = new DefaultHttpClient(connmgr,basicParams);

// It's a lead,be forgiving with timeout
g_client.getParams().setParameter(CoreConnectionPNames.CONNECTION_TIMEOUT,5000);
g_client.getParams().setParameter(CoreConnectionPNames.so_TIMEOUT,5000);
ConnManagerParams.setTimeout(basicParams,2000);

当我调用g_client.execute(postMethod);在我的代码中显示400状态代码意味着我的请求出了问题.如果我在DefaultHttpClient构造函数中传递null

g_client = new DefaultHttpClient(connmgr,null);

客户端正在成功执行,但由于我需要基本参数集maxconnectionperroute,因此它不正确.我在这里粘贴了令人怀疑的代码.请看看并帮助我.我被困在这里.

我使用的是HttpClient 4.0版本.

SchemeRegistry sc = new SchemeRegistry();
sc.register(new Scheme("http",80,PlainSocketFactory.getSocketFactory()));
sc.register(new Scheme("https",443,SSLSocketFactory.getSocketFactory()));

ThreadSafeClientConnManager connmgr = new ThreadSafeClientConnManager(sc);
connmgr.setDefaultMaxPerRoute(5);  /// Alternative approach to yours ///

DefaultHttpClient g_client = new DefaultHttpClient(connmgr);
g_client.getParams().setParameter(CoreConnectionPNames.CONNECTION_TIMEOUT,5000);
g_client.getParams().setParameter(ConnManagerpnames.TIMEOUT,2000);   /// Alternative approach to yours ///