web.xml中的<error-page>标记不会捕获java.lang.Throwable异常

Question

我有一个用servlet和JSP开发的web应用程序.IllegalArgumentException如果我插入错误的参数,我配置我的应用程序抛出一个.然后我以这种方式配置了我的web.xml文件:

<error-page>
    <error-code>404</error-code>
    <location>/error.jsp</location>
</error-page>
<error-page>
    <exception-type>java.lang.Throwable</exception-type>
    <location>/error.jsp</location>
</error-page>

当我上升404 error,然后它工作和呼叫error.jsp,但当我上升java.lang.IllegalArgumentException,然后它不起作用,我有一个blank page而不是error.jsp.为什么？

服务器是Glassfish,日志显示真的IllegalArgumentException上升.

Answer 1

你不应该抓住并压制它,但要放手吧.

即不做:

@Override
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    try {
        doSomethingWhichMayThrowException();
    } catch (IllegalArgumentException e) {
        e.printStackTrace(); // Or something else which totally suppresses the exception.
    }
}

而是放手吧:

@Override
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    doSomethingWhichMayThrowException();
}

或者,如果你真的想要捕获它用于记录等(我宁愿使用过滤器,但是ala),然后重新抛出它:

@Override
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    try {
        doSomethingWhichMayThrowException();
    } catch (IllegalArgumentException e) {
        e.printStackTrace();
        throw e;
    }
}

或者,如果它不是运行时异常,然后重新抛出它包装ServletException,它将被容器自动解包:

@Override
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
    try {
        doSomethingWhichMayThrowException();
    } catch (NotARuntimeException e) {
        throw new ServletException(e);
    }
}

也可以看看:

• 服务器如何确定要使用哪种类型的web.xml错误页面的优先级？
• /sf/ask/681763071/#9746657