Java Restful Service eclipse tomcat HTTP Error 404 [英] Java Restful Service eclipse tomcat HTTP Error 404

问题描述

我尝试跟上java休息服务。所以我找到了一些教程，它们总是以同样的方式解释。但我无法运行：（。

i try to keep up with a java rest service. So i found some tutorials, which explain always the same way. But i cant get this running :(.

我在版本2.5中创建了动态Web项目，在eclipse中创建了Tomcat 7.0。然后我将以下jar加载到WEB-INF / lib

I made Dynamic Web Project in Version 2.5 and Tomcat 7.0 in eclipse. Then i load following jars to WEB-INF/lib

我的项目名是com.freespots.rest。我创建了以下web.xml

My Projectname is com.freespots.rest. I created following web.xml

好了我现在要创建java-class吧？我做到了Java Resources / src / com.freespots.rest.service:

Ok now iam going to create the java-class right? Well i did it Java Resources/src/com.freespots.rest.service:

如果我启动Tomcat并输入url到我的浏览器，如localhost：8080 / com.freespots.rest，Tomcat会显示我的index.html文件
但是，如果我去的URL本地主机：8080 / com.freespots.rest / API /你好只是一个HTTP 404错误：

If i start Tomcat and type url to my browser like localhost:8080/com.freespots.rest the Tomcat shows my index.html file. But if i go to url localhost:8080/com.freespots.rest/api/hello there is just a HTTP 404 Error:

我刚接触Java Webdevelopment并且我无法弄清楚我的问题。我希望有些人可以解释我的错误。在此先感谢。

Well iam new to Java Webdevelopment and i cant figure out my Problem. I hope some guy can explain my mistake. Thanks in advance.

推荐答案

您正在使用旧的（1.x）Jersey配置。在Jersey 2.x中，类名称空间和属性名称已更改。应改用

You're using the old (1.x) Jersey configuration. In Jersey 2.x, the class namescpaces and property names have changed. You should instead use

<servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
<init-param>
    <param-name>jersey.config.server.provider.packages</param-name>
    ...

查看其他部署选项这里

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