How would you make it so that /url works but /url.html or /url.jsp fails? I'd like to setup my pattern to only accept exact matches that have no extension.
Currently in web.xml I use
<servlet-mapping>
<servlet-name>DispatchServlet</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
and on my controllers I use
#RequestMapping(value = "/url", method = RequestMethod.GET)
but that allows any extension to be mapped to my controllers so long as 'url' is contained in the requested path. I would like extensions or incorrectly appended characters to fail. Is there a way to do this without parsing the URL string programmatically for each controller?
Related
I have tested like default, extension, path(start with '/' and end with '/*') these three are working fine but exact match url pattern '/test' showing 404 error.Please refer this is my exact code in web.xml file for configuring dispatcher servlet.
<servlet>
<servlet-name>frontController</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>frontController</servlet-name>
<url-pattern>/test</url-pattern>
</servlet-mapping>
For this url on address bar :http://localhost:8086/MVCFirstApp/test/sugar
I am getting error - 404 page not found for exact match url pattern only.
Please help me out from this, thanks in advance.
Please check few steps before reading the solutions
+ do you have proper controller to handle the requested request(url pattern)
+ do you have proper view returned by the controller
+ and is the base package scanned for controller (annotations)
It would be more convenient if you share the controller code too. But with what I understand with your statement you can try this
add URL patters as
/test/*
Let me explain what this does it will send all the request which comes tru URL /test/ to dispatcher servlet . For example /test/sugar or /test/abc/def/ghi anything, in more simple words any request with URL /test/* is sent to dispatcher servlet which matches and return the proper controller with help of handler mapping .
And make sure you have added the mapping for /test/sugar or /test with proper view in controller . Or if you want sugar to be your value then use #Pathvariable in your controller.
I am using spring 3.1.1. I am trying to HTML encode my incoming request parameters. The call to my JSP page can be made manually by a user passing in a URL from a command line tool or Web browser.
lets assume I have a request as below with request parameter as language
http://localhost:8080/testdomain/createaccount.do?language=eng
I want to HTML encode the 'language' request parameter. I have already set a context-param in web.xml to html encoding.
<context-param>
<param-name>defaultHtmlEscape</param-name>
<param-value>true</param-value>
</context-param>
Questions
Specifying the defaultHtmlEscape within the web.xml does not escape the html elements in the request parameter. Doesn't seem to be working? Any suggestions on how to get this working?
If I call the below inside the controller? Would this even help as the parsing must have already been done?
HtmlUtils.htmlEscape(request.getParameter(language))"
At the specific spring controller for this page, how do I ensure Html encoding before reading the request parameter?
I want to understand options where I can enforce html encoding at the application level and at the page/controller level?
The web descriptor property will apply to Spring tags only. Use:
<form:input path="/my/path" htmlEscape="true" />
Make sure you are protected against XSS when using HtmlUtils.htmlEscape
What you need is just a #RestController
To read a parameter from the URL it is better to use #RequestParam
public #ResponseBody String myRequest(#RequestParam("language") String language) {
System.out.println("Language is: "+ language);
}
Check https://spring.io/guides/gs/rest-service/
From inside my JAX-RS (Jersey) resource I need to get the base URL of the Jersey Servlet that's "publishing" that resource. I tried injecting ServletContext as described here, and then doing a:
return servletContext.getResource("/").toString();
to get the "base" URL of the Jersey Servlet for this resource.
However the above code returns a value like:
jndi:/localhost/jax-rs-tomcat-test/
where I was expecting something more like:
http://localhost:8080/jax-rs-tomcat-test/jax-rs
Where "jax-rs" is what I have in my web.xml:
<servlet-mapping>
<servlet-name>jersey-servlet</servlet-name>
<url-pattern>/jax-rs/*</url-pattern>
</servlet-mapping>
That is, there are four "differences": (a) protocol, (b) single instead of double slash after the protocol, (c) port number and (d) missing URL pattern for triggering the Jersey servlet. So, how do I get:
the base http:// URL of the Jersey servlet
the full URL that triggered a particular #GET or #POST annotated method ?
You're looking for UriInfo. Inject it into your resource using #Context:
#Context
private UriInfo uriInfo;
and then you can call getBaseUri() method:
uriInfo.getBaseUri();
I am back with working in Springs. I used to work in Springs but blindly, didn't understand much. I used to get a lot of errors, very basic ones, and I am getting them again.
My problem is that, I don't know how the configuration of the Spring-MVC work.
What happens when I run the project from my STS?
I am working on the spring template project in STS.
I am getting this when I run the project.
WARN : org.springframework.web.servlet.PageNotFound - No mapping found for HTTP request with URI [/common/] in DispatcherServlet with name 'appServlet'
I am totally fed up and broken.
Just 2 months of break from work, I am back at the starting block.
I don't want to post my code and make the question specific.
I want an answer that explains the way in which the server executes a spring project. Right from the running of an application(basic hello world application) to the display of the home page.
This will be helpful for all the beginners.
I tried searching for such an explanation in the net but I didn't get any proper explanation, but got a lot of basic samples. Those samples are easy to understand but are not explaining the way in which the server goes about.
Note: I am looking for an answer that explains the Springs concept. From the running of an application to the display of a home page. What all happens in this process? Where does the server start with? How does it go about?
Here is the flow initially servlet container loads the web.xml file.In web.xml we will specify that all the requests are handled by the spring FrontController that is DispatcherServlet.
We include it by adding the following code
<servlet>
<servlet-name>dispatcher</servlet-name>
<servletclass>org.springframework.web.servlet.DispatcherServlet</servletclass>
<load-on-startup>2</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>dispatcher</servlet-name>
<url-pattern>*.htm</url-pattern>
</servlet-mapping>
Here it indicate if the url request is of *.htm it is handled by dispatcherServlet then dispatcherServlet load dispatcher-servlet.xml . Where we need to mention the mapping to controller by writing the specific url request such as
<bean name="/insert.htm" class="com.controller.MyController"></bean>
So in bean we mention that for request of /insert.htm it tells the servlet to look in the mentioned class.You need use the Annotation of #RequestMapping above the method for ex
#RequestMapping("/insert.htm")
public ModelAndView insert(HttpServletRequest req,Student student)
{
String name=req.getParameter("name");
int id=Integer.parseInt(req.getParameter("id"));
student.setId(id);
return new ModelAndView("display","Student",student);//It returns a view named display with modelclass name as `Student` and model object student
}
So when a Request url of /insert.htm appears it executes the above method it returns a ModelAndView object nothing but an view.It again goes to dispatcher-servlet.xml and looks for view Resolver the normal code that is to be added is
<bean id="viewResolver"
class="org.springframework.web.servlet.view.InternalResourceViewResolver"
p:prefix="/WEB-INF/jsp/"
p:suffix=".jsp" />
So from this it gets the logical view name and appends the prefix and suffix to it .Finally it displays the content in the view.so it looks for display in view resolver prefixes and suffixes the things and finally returns /WEB-INF/jsp/display.jsp .Which displays the jsp content
You are mapping your Spring servlet only for requests that end with .htm. The request for the root of your application does not end with .htm and so, it does not get picked up by Spring. Edit your web.xml as follows, in order to use Spring for all requests:
<servlet-mapping>
<servlet-name>appServlet</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
Then, use this as the controller:
package com.mkyong.common;
#Controller
public class HomeController {
#RequestMapping(value = "/", method = RequestMethod.GET)
public ModelAndView helloWorld() {
ModelAndView model = new ModelAndView("index");
model.addObject("msg", "hello world");
return model;
}
}
The controller intercepts the requests for the context root of the application, adds the msg attribute to the model and redirects to the index view.
So, you need to add the index.jsp file in the /WEB-INF/views/ directory. Inside your jsp, you will be able to use the value of the msg attribute.
From what every you have posted you do no have a request mapping for the url /common/.
You will have to create another request mapping function like the one below in your controller class and create a view file also.
#RequestMapping(value = "/common/", method = RequestMethod.GET)
public ModelAndView common(HttpServletRequest request,
HttpServletResponse response) {
ModelAndView model = new ModelAndView("common");
model.addObject("msg", "hello world");
return model;
}
Fresh off my last adventure, now I'm trying to map more complex URLs and going nuts trying to make it work the way Spring's documentation suggests it should.
Again, the tools are:
Java 1.6
Spring 3.2 MVC
Tomcat 7
What I'm trying to do is match URLs of the form foo/bar/id where id is an integer. The way it seems like I should do it is to annotate my controller method like this:
#RequestMapping("/foo/bar/{id}")
And then have this in web.xml:
<url-filter>/foo/*</url-filter>
Or this:
<url-filter>/foo/bar/*</url-filter>
And then after deploying to Tomcat, I should be able to access /mycontext/foo/bar/id. But that doesn't work.
For completeness, here several variations and results:
Method mapping: /foo, url-filter: /foo, result: /mycontext/foo works.
Method mapping: /foo/*, url-filter: /foo/*, result: successful mapping of method to /foo/* reported at deployment, but accessing /mycontext/foo/bar fails.
Method mapping: /foo/*, url-filter: /foo/bar, result: mapped at deployment, but accessing /mycontext/foo/bar fails.
Method mapping: /foo/bar, url-filter: /foo/bar, result: /mycontext/foo/bar works.
Method mapping: /foo/bar/*, url-filter: /foo/bar/*, result: mapped at deployment, but accessing /mycontext/foo/bar/(anything) fails.
Method mapping: /foo/bar/{id}, url-filter: /foo/bar/*, result: mapped at deployment, but accessing /mycontext/foo/bar/(anything) fails.
All of the failures come with error messages from the DispatcherServlet for mycontext that no mapping was found, even though all of them reported success in setting up mapping at deployment time. Since I'm getting the error from the right DispatcherServlet, that suggests my url-filter settings are fine. But the message about successful mapping at deployment references whatever is in the #RequestMapping annotation, so I don't know what to make of Spring first saying it's fine and then later saying it doesn't match.
Is there something I've failed to understand about wildcards here?
If you have #RequestMapping("/foo/bar/{id}") this is mapped after combining with the DispatcherServlet's url-pattern match. Consider a case for eg, where the url-pattern is as follows:
<servlet-mapping>
<servlet-name>spring</servlet-name>
<url-pattern>/dispatcher</url-pattern>
</servlet-mapping>
In this case the DispatcherServlet will map the method only if the call from the client is to : /dispatcher/foo/bar/1
So if you want say #RequestMapping to respond to http://<server>/<context>/foo/bar/1 say, just put your url-pattern for DispatcherServlet as / instead:
<servlet-mapping>
<servlet-name>spring</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>