使用servlet过滤器从发布的数据中删除表单参数 [英] Use servlet filter to remove a form parameter from posted data
问题描述
供应商一直在通过HTTPS在名为XMLContent的表单变量中将XML数据发布到我的Coldfusion应用程序服务器。我最近转移到更新版本的应用程序服务器，这些请求丢失了500个服务器错误。它抛出了错误，因为第二个表单参数的内容没有正确地进行urlencoded，但我还是不需要那个参数。 （我联系了供应商来解决这个问题，但是他们强迫我付钱修理他们的错误，所以如果可能的话，我希望自己修理它。）
A vendor has been posting XML data over HTTPS within a form variable named XMLContent to my Coldfusion application server. I recently moved to a more recent version of the application server and those requests are throwing 500 server errors. It is throwing the error because a second form parameter's content is not properly urlencoded, but I don't need that parameter anyway. (I contacted the vendor to fix this but they are forcing me to pay to fix their mistake so I'm looking to fix it myself if possible.)
我怎么办？利用servlet过滤器删除除了名为的形式参数之外的所有参数：XMLContent
我尝试过各种尝试来显式删除有问题的参数TContent，但它永远不会被删除。
How would I utilize a servlet filter to remove all but the form parameter named: XMLContent I have tried various attempts to explicitly remove the offending parameter "TContent" but it never gets removed.
收到的数据片段：
XMLContent=%3C%3Fxml+version%3D%221.0%22+encoding%3D%22UTF-8%22%3F%3E%0A%3CCheck+xmlns%3D%22http .........&TContent=<!--?xml version="1.0" encoding="UTF-8"?--><check xmlns="http...........
我试过的代码：
import java.io.IOException;
import java.util.Collections;
import java.util.Enumeration;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import javax.servlet.Filter;
import javax.servlet.FilterChain;
import javax.servlet.FilterConfig;
import javax.servlet.ServletException;
import javax.servlet.ServletRequest;
import javax.servlet.ServletResponse;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletRequestWrapper;
import java.util.*;
public class MultipartFilter implements Filter {
// Init ----------------------------------------------------------------
public FilterConfig filterConfig;
// Actions -------------------------------------------------------------
public void init(FilterConfig filterConfig) throws ServletException {
this.filterConfig = filterConfig;
}
/**
* Check the type request and if it is a HttpServletRequest, then parse the request.
* @throws ServletException If parsing of the given HttpServletRequest fails.
* @see javax.servlet.Filter#doFilter(
* javax.servlet.ServletRequest, javax.servlet.ServletResponse, javax.servlet.FilterChain)
*/
public void doFilter(ServletRequest request, ServletResponse response, FilterChain chain)
throws ServletException, IOException
{
// Check type request.
if (request instanceof HttpServletRequest) {
// Cast back to HttpServletRequest.
HttpServletRequest httpRequest = (HttpServletRequest) request;
// Parse HttpServletRequest.
HttpServletRequest parsedRequest = parseRequest(httpRequest);
// Continue with filter chain.
chain.doFilter(parsedRequest, response);
} else {
// Not a HttpServletRequest.
chain.doFilter(request, response);
}
}
/**
* @see javax.servlet.Filter#destroy()
*/
public void destroy() {
this.filterConfig = null;
}
private HttpServletRequest parseRequest(HttpServletRequest request) throws ServletException {
// Prepare the request parameter map.
Map<String, String[]> parameterMap = new HashMap<String, String[]>();
// Loop through form parameters.
Enumeration<String> parameterNames = request.getParameterNames();
while (parameterNames.hasMoreElements()) {
String paramName = parameterNames.nextElement();
String[] paramValues = request.getParameterValues(paramName);
// Add just the XMLContent form parameter
if (paramName.equalsIgnoreCase("xmlcontent")) {
parameterMap.put(paramName, new String[] { paramValues[0] });
}
}
// Wrap the request with the parameter map which we just created and return it.
return wrapRequest(request, parameterMap);
}
// Utility (may be refactored to public utility class) ---------------
/**
* Wrap the given HttpServletRequest with the given parameterMap.
* @param request The HttpServletRequest of which the given parameterMap have to be wrapped in.
* @param parameterMap The parameterMap to be wrapped in the given HttpServletRequest.
* @return The HttpServletRequest with the parameterMap wrapped in.
*/
private static HttpServletRequest wrapRequest(
HttpServletRequest request, final Map<String, String[]> parameterMap)
{
return new HttpServletRequestWrapper(request) {
public Map<String, String[]> getParameterMap() {
return parameterMap;
}
public String[] getParameterValues(String name) {
return parameterMap.get(name);
}
public String getParameter(String name) {
String[] params = getParameterValues(name);
return params != null && params.length > 0 ? params[0] : null;
}
public Enumeration<String> getParameterNames() {
return Collections.enumeration(parameterMap.keySet());
}
};
}
}
推荐答案
我们在处理语言环境时，每天都要面对这些情况。如果用户区域设置与浏览器区域设置不同，我们必须更新请求。但它不可变。
We face these situations every day while handling locale. If user locale is different than browser locale, we have to update the request. But its not mutable.
解决方案：创建一个请求包装类。将现有的请求参数（您想要的参数）复制到其中，并在filterchain.doFilter（）中传递此包装类。
Solution: create a request wrapper class. Copy existing request parameters (the ones you want) into it and pass this wrapper class in filterchain.doFilter()
示例包装类：
public class WrappedRequest extends HttpServletRequestWrapper
{
Map<String, String[]> parameterMap;
public WrappedRequest(final HttpServletRequest request)
{
//build your param Map here with required values
}
@Override
public Map<String, String[]> getParameterMap()
{
//return local param map
}
//override other methods if needed.
}
现在在您的过滤器代码中，请执行以下操作。
Now in your filter code, do following.
wrapRequest = new WrappedRequest(hreq);
filterChain.doFilter(wrapRequest, servletResponse);
希望它能解决您的问题。
Hopefully it will solve your problem.
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