Lets say I hit
http://localhost/webapp/wcs/stores/servlet/en/marksandspencer/l/women/dresses/party-and-cocktail-dresses
and this internally redirects me to custom 404.jsp page, But URL remain same in address bar.
I tried this code - <%= request.getAttribute("javax.servlet.forward.request_uri") %>; and it's returning me the path of 404.jsp
How can I get the entered URL which is there in address bar?
Use request.getAttribute("javax.servlet.error.request_uri") to get URI of requested page that not found (404 error). Check this: https://tomcat.apache.org/tomcat-7.0-doc/servletapi/constant-values.html
When error raised (because of some reason such as page not found (404), Internal Server Error (500), ...), the servlet engine will FORWARD the request to corresponding error page (configured in web.xml) using ERROR dispatcher type, NOT FORWARD dispatcher type so that is the reason we must use javax.servlet.error.request_uri, NOT use javax.servlet.forward.request_uri
I think you were close. javax.servlet.forward.request_uri is for normal forwarding, but for 404, you need javax.servlet.error.request_uri.
You can use :
String url = request.getRequestURL().toString();
but this doesn't hold Query String. So, to get query string, you may call
request.getQueryString()
You can do this to get the whole URL including parameters.
request.getRequestURL()+""+request.getQueryString();
use request.getHeader("Referer").
referer gives a url from where you redirected.
Related
I need help on the issue where we are appending special character like % with URL.
Suppose my application URL is "http://www.google.com/".
we have our custom error page which will come when server will not find URL address.
scenario 1: hitting "http://www.google.com/sdfkdjkfj :---redirecting on custom error page which is correct.
Scenario 2: hitting "http://www.google.com/% :--Instead of redirecting on my own custom page it is showing message "this page is not working HTTP ERROR 400".
we are using below code in web.xml to handle the page not found exception.
<error-page>
<error-code>404</error-code>
<location>/ErrorPage.jsp</location>
.
Please help on scenario 2 .
If you want to include a % character in a URL, it needs to be percent-encoded; e.g. http://www.google.com/%25. (That URL still may not be recognized, but at it is syntactically well-formed.)
Error code 400 means "Bad request", the server cannot understand this request. Generally you should avoid passing special character "%" in your URL like in your example.
However if you want to pass this character and preserve its meaning, try to encode it: https://www.w3schools.com/tags/ref_urlencode.asp. When you try http://www.google.com/%25 - you will have 404 error.
I have a URL shortner that should sendRedirect(URL) to URLs specified by users.
Sometimes URL contain curly braces like this: http://example.com?someparam={something}.
Instead of sending response 302 to client browser, my Spring MVC app at Tomcat server gives error 404 with no text.
Apparently it's some sort of URL variable evaluation taking place, can I disable it? I could not find docs regarding this feature.
I know this is an old question but I think the OP was looking for a way to prevent Spring from doing variable replacement in redirect URL
I faced the exact same issue and the fix was using RedirectView
and in RedirectView you can set setExpandUriTemplateVariables(false)
that made it redirect to the url given exactly without Spring trying to replace anything in it
here is how the code looks like
RedirectView redirect = new RedirectView(redirectUrl);
redirect.setExpandUriTemplateVariables(false);
return new ModelAndView(redirect);
Hope that helps
This is not valid Google search URL http://google.com/{something}. It should have been https://www.google.ca/search?q=http{302}
Emphasis is on search?q. After domain name you have specify your service name and then query string if you want to pass some inputs.
When you do http://google.com/{something} then you really do not have any resource or service as {something} so 404 is the expected output.
HTTP 302 is for redirection, I am not sure why you were expecting redirection.
URL encoding will also not help because issue is related to resource/service, if it is not present then you will get 404. URL encoding is not meant to solve problem related to 404.
Reading below link , I could note that "doGet() allows bookmarks".
http://www.developersbook.com/servlets/interview-questions/servlets-interview-questions-faqs.php : search "It allows bookmarks"
Can anyone tell how and what is the use of it ?
All the parameters of GET request are contained in the url so when you are requesting for a resource using GET request, it can be formed using request URL itself.
Consider an example www.somesite.com.somePage.jsp. This generates a GET request because we are asking for a resource somePage.jsp.
If you are asking for a resource, then it is the GET request.
GET requests are used to retrieve data.
any GET request calls the doGet() method of servlet
GET requests are idempotent, i.e. calling the same resource again and again do not cause any side effects to the resources.
Hence, a GET request can have bookmarks
EDIT :-
As suggested by Jerry Andrews, POST methods do not have the query data unlike GET requests to form the resource properly with the help of only url. Hence they are not bookmarked.
It means that If you bookmark the URL of the servlet that has doGet() implemented, you could always get the same page again when you re-visit. This is very common when you have searches, link for products, news, etc.
I need to populate some request with data and redirect back. Is there Spring RedirectAttributes analog for Java EE? I have searched and found 2 solutions, but they also have limitation:
Response.sendRedirect()
In this case I will lost all destroys request attributes. I can use Session attributes but in this case I need some mechanism that can determine when redirect comes in or when there is no redirect and data must be removed.
getRequestDispatcher(String path).forward(request,response)
The problem with path - I need to send redirect to URL not to give something jsp or Servlet by name. Is there any way to "convert" redirect URL to path? For example how I can go forward to
"http://localhost:8080/WebAppname/"?
You can use the sendRedirect and pass the parameters as part of the query string. So what you would be redirecting can be something like below
http://localhost:8080/WebAppname/myRedirect.action?param1=value1¶m2=value2
In my servlet i'm trying to response.sendRedirect() a request to another servlet (which is deployed via different .war). I understood that sendRedirect() makes a new request while forwarding, but looks like there is no mechanism to change new request's method type. Basically i'm trying to see if there is an option to choose Get or Post while calling sendRedirect.
Please suggest me if there is a way to change the request type or if not i'm trying to understand why it is not allowed.
response.sendRedirect is always a GET [also in broad term , redirect always GET s]. You can forward for your choice
See Also
java-servlet-difference-between-send-redirect-and-forward-in-servlets
URL Redirection
response.sendRedirect will always be a get as Jigar mentioned.
You can post to the target by having a hidden form with method post and submitting it with javascript when the page is loaded.