深入Ajax代理的Java Servlet的實現詳解

代碼如下所示： 複製代碼 代碼如下:import java.io.IOException;
import java.io.InputStream;
import java.net.URL;
import javax.servlet.ServletConfig;
import javax.servlet.ServletException;
import javax.servlet.ServletOutputStream;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
/**
* Take any request and proxy it to the given REDIRECT_BASE.
* For example, if this servlet lives at
*
* http://foo.com/forward
*
* and is inititialized with the REDIRECT_BASE
*
* http://bar.com/some/path
*
* then a GET request like
*
* http://foo.com/forward?quux=mumbley
*
* will return the results of a GET from
*
* http://bar.com/some/path?quux=mumbley
*
* This is not robust and generalized; it's simple and quick.
*
* @author jdf
*
*/
public class ProxyServlet extends HttpServlet
{
private final static String COPYRIGHT = com.ibm.dogear.Copyright.SHORT;
public static final String REDIRECT_BASE = "com.ibm.bl.servlet.RedirectServlet.redirect_base";
private String redirectBase;

@Override
public void init(ServletConfig config) throws ServletException
{
super.init(config);
redirectBase = getRequiredParam(REDIRECT_BASE);
}
@Override
protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws IOException
{
String queryString = req.getQueryString();
URL url = new URL(redirectBase + (queryString != null ? "?" + queryString : ""));
copyInputStreamToOutputStream(url.openStream(), resp.getOutputStream());
}
private void copyInputStreamToOutputStream(InputStream in, ServletOutputStream out)
throws IOException
{
try
{
try
{
byte[] buffer = new byte[1024];
int n;
while ((n = in.read(buffer)) != -1)
out.write(buffer, 0, n);
}
finally
{
out.close();
}
}
finally
{
in.close();
}
}
protected String getRequiredParam(String param) throws ServletException
{
String result = getServletConfig().getInitParameter(param);
if (result == null) {
throw new ServletException(getClass() + " requires " + param + " param");
}
return result;
}
}