我正在尝试将某个字符串附加到发送回客户端的每个响应中。为此,我使用如下所示的过滤器:
public class OutFilter implements Filter {
@Override
public void doFilter(ServletRequest req, ServletResponse resp,
FilterChain filterChain) throws IOException, ServletException {
PrintWriter out = resp.getWriter();
CharResponseWrapper wrapper = new CharResponseWrapper(
(HttpServletResponse) resp);
filterChain.doFilter(request, wrapper);
out.print(wrapper.toString());
out.print("ADDITIONAL CONTENT".toString());
}
}
我知道在 doFilter() 方法通过后我无法更改 ServletResponse 对象,所以这就是我使用包装器进行 HTTP-Response 的原因:
public class CharResponseWrapper extends HttpServletResponseWrapper {
private CharArrayWriter output;
private ServletOutputStream stream;
public String toString() {
return output.toString();
}
public CharResponseWrapper(HttpServletResponse response) throws IOException {
super(response);
output = new CharArrayWriter();
stream = new ServletOutputStream() {
@Override
public void write(int b) throws IOException {
// TODO Auto-generated method stub
}
};
}
public PrintWriter getWriter() {
return new PrintWriter(output);
}
public ServletOutputStream getOutputStream() {
return stream;
}
}
我的 web.xml 看起来像这样:
<filter>
<filter-name>OutFilter</filter-name>
<filter-class>com.foo.OutFilter</filter-class>
</filter>
<filter-mapping>
<filter-name>OutFilter</filter-name>
<url-pattern>/1.html</url-pattern>
<url-pattern>/2.html</url-pattern>
<url-pattern>/3.html</url-pattern>
<url-pattern>/statistics</url-pattern>
<dispatcher>FORWARD</dispatcher>
<dispatcher>REQUEST</dispatcher>
</filter-mapping>
统计 servlet 如下所示:
public class Statistics extends HttpServlet {
@Override
protected void doGet(HttpServletRequest req, HttpServletResponse resp)
throws ServletException, IOException {
resp.getWriter().print("<b>Hello, World</b>");
}
}
关键是在所有访问文件的 HTTP 响应末尾附加“附加内容”,因此 1.html、2.html、3.html 以及 /statistics(服务于 Statistics Servlet)。
访问 servlet 工作正常 - 我得到了 html 内容,然后是“附加内容”字符串。但是,访问任何一个静态 html 文件都会产生以下错误:
java.lang.IllegalStateException: NO CONTENT
我究竟做错了什么?
感谢所有帮助。