处理未映射的休息路径 [英] Handling un-mapped Rest path

问题描述

这是我当前的Guice配置:

This is my current Guice configuration:

public class MyServletModule extends ServletModule {
    @Override
    protected void configureServlets() {
        bind(MyRest.class);
        serveRegex(".+(?<!\\.(html|css|png|jpg))")
               .with(HttpServletDispatcher.class);
    }
}

但是我希望我的Rest资源只能以http://127.0.0.1:8888/{hashcode_or_filename}的形式访问，并且是唯一接受和处理的形式(嗯，加上下面的/create方法).

However I want that my Rest resource is only access in form of http://127.0.0.1:8888/{hashcode_or_filename} and the only form accepted and processed (well, plus the /create method below).

现在，我可以在此路径模式中正确处理哈希码和文件名.

Right now, I can deal with hashcode and filename properly in this path pattern.

但是我不确定在客户端请求未映射路径的情况下，如何处理下面的种类或情况，在我的情况下会返回此路径:

However I am not sure how to deal the kind or scenario below, where the client is requesting path that is not mapped, which returns this in my case:

找不到相对资源:/examples/foo的完整路径: http://127.0.0.1:8888/examples/foo

Could not find resource for relative : /examples/foo of full path: http://127.0.0.1:8888/examples/foo

或

找不到相对资源:/examples/bar/foo已满 路径:http://127.0.0.1:8888/examples/bar/foo

Could not find resource for relative : /examples/bar/foo of full path: http://127.0.0.1:8888/examples/bar/foo

我需要的是能够处理未映射的路径，以便我可以返回错误的HTML页面或其他内容，而不在浏览器中显示这些错误文本.

What I need is to be able to be able to handle unmapped paths so I can return a error HTML page or something and not show these error text in the browser.

如果请求是:http://127.0.0.1:8888/，我也需要自动转发到http://127.0.0.1:8888/index.html.现在，我必须手动将index.html放在尾部.

Also if the request is: http://127.0.0.1:8888/ I need to forward to http://127.0.0.1:8888/index.html automatically. As right now I have to manually put the index.html in the tail.

我的Resteasy资源仅通过以下方式进行配置或接线:

My Resteasy resource is configure or wired with just:

@Singleton
@Path("/")
public class MyRest {
    @GET
    @Path({hashcode})
    public Response getSomething(...){}

    @POST
    @Path("create")
    public Response createSomething(...){}
}

推荐答案

最简单的方法是注册过滤器以处理错误代码不是200(确定)的响应.或在您的web.xml中添加如下内容:

Easiest way is to register filter to handle responses with error code other that 200 (OK). Or add to your web.xml something like this:

<error-page>
    <error-code>404</error-code>
    <location>/ErrorPage.jsp</location>
</error-page> 

如果请求是:http://127.0.0.1:8888/我还​​需要转发给 http://127.0.0.1:8888/index.html自动.现在，我必须 手动将index.html放在尾部.

Also if the request is: http://127.0.0.1:8888/ I need to forward to http://127.0.0.1:8888/index.html automatically. As right now I have to manually put the index.html in the tail.

您可以使用此模块 http://tuckey.org/urlrewrite/

WEB-INF/web.xml

<filter>
    <filter-name>UrlRewriteFilter</filter-name>
    <filter-class>org.tuckey.web.filters.urlrewrite.UrlRewriteFilter</filter-class>
    <init-param>
        <param-name>confPath</param-name>
        <param-value>/WEB-INF/urlrewrite.xml</param-value>
    </init-param>
    <!--...omitted...-->
</filter>

<filter-mapping>
    <filter-name>UrlRewriteFilter</filter-name>
    <url-pattern>/*</url-pattern>
    <dispatcher>REQUEST</dispatcher>
    <dispatcher>FORWARD</dispatcher>
</filter-mapping>

WEB-INF/urlrewrite.xml

<?xml version="1.0" encoding="utf-8"?>
<!DOCTYPE urlrewrite
        PUBLIC "-//tuckey.org//DTD UrlRewrite 4.0//EN"
        "http://www.tuckey.org/res/dtds/urlrewrite4.0.dtd">
<urlrewrite>
    <rule match-type="regex">
        <from>^/$</from>
        <to type="redirect">/index.html</to>
    </rule>
</urlrewrite>

这篇关于处理未映射的休息路径的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！