现在假设我触发一些 URL 说
http://shoaib:8080/userAuthentication?authenticate=false
现在到了Filter的doFilter方法
public void doFilter(request,response,filterChain){
boolean authenticate=request.getParameter("authenticate");
.
. //some logic
.
.
filterChain.doFilter(request,response); //this request should not have query String.
}
将传递给 filterChain 的 doFilter()
的请求不应该有 queryString。我怎样才能做到这一点?
最佳答案
我在考虑覆盖 Filter 方法:
@Override
public void doFilter(ServletRequest req, ServletResponse res, FilterChain chain)
throws ServletException, IOException {
HttpServletRequest request = (HttpServletRequest) req;
String requestURI = request.getRequestURI();
if (requestURI.startsWith(YOUR_CHECK_PATH)) {
String url = ((HttpServletRequest)request).getRequestURL().toString();
String queryString = ((HttpServletRequest)request).getQueryString();
req.getRequestDispatcher(url).forward(req, res);
} else {
chain.doFilter(req, res);
}
}
关于java - 在 ServletRequest 到达 Servlet/Filter 后从中删除查询字符串,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/22499516/