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[英]How to use a design pattern to arrange the file structure in a java+spring application
[英]How do I correctly use @PathVariable in Java+Spring?
我有一个控制器可以很好地处理其他请求映射,但是我很难@PathVariable
与@PathVariable
注释一起使用。
在此示例中 (请参见“ URI模板模式”),它具有以下示例:
@RequestMapping(value="/owners/{ownerId}", method=RequestMethod.GET)
public String findOwner(@PathVariable String ownerId, Model model) {
Owner owner = ownerService.findOwner(ownerId);
model.addAttribute("owner", owner);
return "displayOwner";
}
我的代码如下所示:
@RequestMapping( value="/ViewReport/json", method = RequestMethod.GET)
public ModelAndView TestJson(final Model model){
model.addAttribute("fnord", "{\"hello\":\"awesome\"}");
return new ModelAndView("reportViewJson");
}
@RequestMapping( value="/ViewReport/jayson/{foobar}", method = RequestMethod.GET)
public ModelAndView TestJayz(final Model model, @PathVariable String foobar){
model.addAttribute("fnord", "{\"hello\":\""+foobar+"\"}");
return new ModelAndView("reportViewJson");
}
然后,我有一个包含以下内容的web.xml
文件:
<servlet-mapping>
<servlet-name>autoreport</servlet-name>
<url-pattern>/ViewReport/jayson/*</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>autoreport</servlet-name>
<url-pattern>/ViewReport/json</url-pattern>
</servlet-mapping>
现在,当我导航到/ViewReport/json
,便得到了我所期望的JSON {"hello": "awesome"}
。 但是,当我导航到/ViewReport/jayson/42
我得到了
HTTP Status 404 -
type Status report
message
description The requested resource () is not available.
Apache Tomcat/7.0.29
我对PathVariable
使用有什么错,我需要怎么做才能解决?
当我的服务器启动时,它弹出以下行:
[ INFO] 14:15(AbstractUrlHandlerMapping.java:registerHandler:315)
Mapped URL path [/ViewReport/json.*] onto handler 'reportViewController'
[DEBUG] 14:15(AbstractBeanFactory.java:doGetBean:246)
Returning cached instance of singleton bean 'reportViewController'
[ INFO] 14:15(AbstractUrlHandlerMapping.java:registerHandler:315)
Mapped URL path [/ViewReport/json/] onto handler 'reportViewController'
[DEBUG] 14:15(AbstractBeanFactory.java:doGetBean:246)
Returning cached instance of singleton bean 'reportViewController'
[ INFO] 14:15(AbstractUrlHandlerMapping.java:registerHandler:315)
Mapped URL path [/ViewReport/jayson/{foobar}] onto handler 'reportViewController'
当我发出请求时,我得到了:
[DEBUG] 20:45(DispatcherServlet.java:doService:823)
DispatcherServlet with name 'autoreport' processing GET request for [/ViewReport/jayson/fnord]
[ WARN] 20:45(DispatcherServlet.java:noHandlerFound:1108)
No mapping found for HTTP request with URI [/ViewReport/jayson/fnord] in DispatcherServlet with name 'autoreport'
春天有魔力; 参数顺序很重要。 尝试反转处理程序参数的顺序。 代替这个:
public ModelAndView TestJayz(final Model model, @PathVariable String foobar)
尝试这个:
public ModelAndView TestJayz(@PathVariable String foobar, final Model model)
采取2:
尝试将web.xml中的映射更改为更通用的映射。 例如:
<servlet-mapping>
<servlet-name>autoreport</servlet-name>
<url-pattern>/ViewReport/*</url-pattern>
</servlet-mapping>
我在测试web.xml中使用了/spring/*
,在控制器中使用了@RequestMapping(“ / blammy / {value}”),然后将请求发送至“ {WEBROOT} / spring / blammy / testvalue”,其中“ {WEBROOT} ”是我的测试应用的网络根目录。
请求映射相对于调度程序Servlet映射的相对关系(默认情况下,请参阅DefaultAnnotationHandlerMapping
和AnnotationMethodHandlerAdapter
alwaysUseFullPath
属性以获取更多信息)。
如果您请求资源/ViewReport/jayson/fnord
spring mvc正在寻找映射到fnord
的处理程序。
您需要请求/ViewReport/jayson/ViewReport/jayson/fnord
您是否尝试过@PathVariable(value =“ foobar”)字符串foobar ?
@RequestMapping( value="/ViewReport/jayson/{foobar}", method = RequestMethod.GET)
public ModelAndView TestJayz(final Model model, @PathVariable(value="foobar") String foobar){
model.addAttribute("fnord", "{\"hello\":\""+foobar+"\"}");
return new ModelAndView("reportViewJson");
}
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