如果我有一个抛出异常的servlet,我如何处理异常以便jsp调用页面,只重新加载正文?
我有一个执行此操作的servlet:
protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
try {
.....
} catch (Throwable th) {
throw new ServletException(th);
}
}
我的error.jsp以这种方式制作:
<%@ page language="java" contentType="text/html; charset=UTF-8"
pageEncoding="UTF-8"%>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<%@ taglib uri="http://struts.apache.org/tags-tiles" prefix="tiles" %>
<tiles:insert page="/include/baseLayout.jsp" flush="true">
<tiles:put name="header" value="/include/header.jsp" />
<tiles:put name="menu" value="/include/menu.jsp" />
<tiles:put name="body" value="/include/body.jsp" />
<tiles:put name="footer" value="/include/footer.jsp" />
</tiles:insert>
tiles.defs.xml以这种方式完成:
...
<definition name="error" extends="baseLayout" controllerClass="it.present.mininterno.prum.localizzazione.exception.ExceptionHandler">
<put name="body" value="/include/body.jsp" />
</definition>
如何运行仅对页面正文部分进行充值 错误信息?该程序是用struts编写的,当我有一个例外时 行动可以做到这一点。正常的servlet没有!!