EclipseJavaRESTAPI方法在发布后抛出404NotFound
我将我的后端托管在Google App Engine上,Rest API在localhost上运行良好。方法url如下所示:
http://127.0.0.1:8888/rest/plans/getplans/242353
这在本地主机上工作,并正确地返回JSON响应。这是一个GET方法,242353是我传递给该方法的参数
但当我将其发布到我的谷歌云时,在我尝试使用以下URL在浏览器上调用它之后:
http://2-dot-MY_APP_DOMAIN.appspot.com/rest/plans/getplans/242353
我在浏览器上遇到此错误:
Error: Not Found
The requested URL /rest/plans/getplans/242353 was not found on this server.
网络。xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee
http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
version="2.5"
xmlns="http://java.sun.com/xml/ns/javaee">
<servlet>
<servlet-name>Jersey Web Application</servlet-name>
<servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>com.dinukapj.socialapp.api</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Jersey Web Application</servlet-name>
<url-pattern>/resources/*</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>Jersey Web Application</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
</web-app>
为什么会发生这种情况?我是否需要向我的网站添加任何其他信息。xml
# 1 楼答案
试着点击你网站上定义的/resources/plans/getplans/242353。xml