Is there a way do not enter a specifc url mapping (vaadin.urlMapping)? But I also want to still use swagger for example (ip:8080/swagger-ui/index.html)? It would be great to exclude some mappings from vaadin.
Thanks for your help!
Best Regards, Thomas
Put the URL/Part of URL/IP in the properties file.
Put the URL/Part of URL/IP in the web.xml file.
In some cases you want the remote URL to be different than the local URL (or remote SSL URL to be different than the local URL).
In that case, I placed both IP and domain on to web.xml in two different contexts.
final String serverUrl = new URL("" + ((VaadinServletRequest)VaadinService.getCurrentRequest())
.getHttpServletRequest().getRequestURL()).getAuthority();
final String serverProtocol =
((VaadinServletRequest)VaadinService.getCurrentRequest()).getHttpServletRequest().getProtocol();
final boolean serverSecure =
((VaadinServletRequest)VaadinService.getCurrentRequest()).getHttpServletRequest().isSecure();
if(!serverSecure){
UI.getCurrent().getSession().setAttribute(
ResourceProperty.configBundle.getString("APPLICATION_URL"),
"http:" + serverUrl);}
else{
UI.getCurrent().getSession().setAttribute(
ResourceProperty.configBundle.getString("APPLICATION_URL"),
"https:" + serverUrl);}
Related
How to exclude particular format of URL from the filter?
Is there a standard method or alternative solution ?
This isn't possible via <url-pattern> configuration.
You've basically 2 options:
Make <url-pattern> more specific. E.g. move all those resources wich really need to be filtered in a specific folder like /app and set the <url-pattern> to /app/*. Then you can just put the excluded page outside that folder.
Check for the request URI in the filter and just continue the chain if it represents the excluded page.
E.g.
String loginURL = request.getContextPath() + "/login.jsp";
if (request.getRequestURI().equals(loginURL)) {
chain.doFilter(request, response);
}
else {
// ...
}
On my website each page has links that are created in the codebehind, where the links are the current URL with one query parameter changed. To do this, I've been using this method (this specific example is for the pagination):
var queryValues = HttpUtility.ParseQueryString(HttpContext.Current.Request.QueryString.ToString());
queryValues.Set("page", num);
string url = HttpContext.Current.Request.Url.AbsolutePath;
string updatedQueryString = "?" + queryValues.ToString();
string newUrl = url + updatedQueryString;
return newUrl;
This worked on my local version fine. However, when I created each page in Ektron and added a manual alias, the URLs generated still went to the file location in the solution. For example, my original page was /WebAssets/Templates/EventListView.aspx. I created the page in Ektron as /Alumni/Events/List. I can go to /Alumni/Events/List, but then when I click on a page button the page that loads is /WebAssets/Templates/EventListView.aspx?page=2 instead of /Alumni/Events/List/?page=2
I found one solution:
var rawUrl = HttpContext.Current.Request.RawUrl;
var url = rawUrl.Split('?')[0];
string newUrl = url + updatedQueryString;
Use the QuickLink property of the primary contentblock for /Alumni/Events/List, this will be the alias use want to use for your page links or for redirects to the same page. This is probably ContentData.QuickLink if you're already loading the ContentData at some point in the code.
Notes:
Aliasing may remove your "page" querystring parameter by default, to resolve this issue, edit your alias in the Workarea to have a "Query String Action" of "Append".
Make sure you preprend a "/" to the QuickLink value (if it's not absolute and not prepended already) if using it on the frontend, otherwise your links will bring you to something like /Alumni/Events/List/Alumni/Events/List?page=2, which is no good.
I have a link to <a href='/ViewReport'> on my local host that works fine, but on the server the whole site is in a folder "serverfolder", so the link becomes http://somesite/serverfolder/ViewReport, which isn't a valid url. I have seen how to use ~ to access the root directory for files, but not how to do this with url paths. I want to use the same link for both local and remote deployment. How would I achieve this? Thank you!
Do this:
var urlHelper = new UrlHelper(Request.RequestContext);
string url = Request.Url.GetLeftPart(UriPartial.Authority)
+ urlHelper.Action("ViewReport",
new { userId = UserName, reportId = PI.ElementAt(i).TempUserID });
Or, if you prefer not to use the UrlHelper.Action, you do it like this:
string url = Request.Url.GetLeftPart(UriPartial.Authority) + "/ViewReport...";
In Grails, I'm trying to make it so that any requests to /images/* actually goes to another website on another host.
I know how to do it in Apache with Mod Rewrite, but how can I achieve this with UrlMappings?
I want
/images/* to go to http://somedomain/images/*
You can't directly. You'll have to map it to a controller which will in turn redirect to the desired location. Something like:
"/images/$urlTail**" (controller: "image", action: "external")
And then in the external method of the ImageController:
def actualUri = request.forwardURI.replace("/images/", "")
redirect "http://example.com/" + actualUri;
Another workaround I've found:
String url = <your-url>
java.net.URI uri = new java.net.URI(url)
return Response.seeOther(uri).build()
I'd like to create URLs based on the URL used by the client for the active request. Is there anything smarter than taking the current HttpServletRequest object and it's getParameter...() methods to rebuilt the complete URL including (and only) it's GET parameters.
Clarification: If possible I want to resign from using a HttpServletRequest object.
Well there are two methods to access this data easier, but the interface doesn't offer the possibility to get the whole URL with one call. You have to build it manually:
public static String makeUrl(HttpServletRequest request)
{
return request.getRequestURL().toString() + "?" + request.getQueryString();
}
I don't know about a way to do this with any Spring MVC facilities.
If you want to access the current Request without passing it everywhere you will have to add a listener in the web.xml:
<listener>
<listener-class>org.springframework.web.context.request.RequestContextListener</listener-class>
</listener>
And then use this to get the request bound to the current Thread:
((ServletRequestAttributes) RequestContextHolder.currentRequestAttributes()).getRequest()
Instead of using RequestContextHolder directly, you can also use ServletUriComponentsBuilder and its static methods:
ServletUriComponentsBuilder.fromCurrentContextPath()
ServletUriComponentsBuilder.fromCurrentServletMapping()
ServletUriComponentsBuilder.fromCurrentRequestUri()
ServletUriComponentsBuilder.fromCurrentRequest()
They use RequestContextHolder under the hood, but provide additional flexibility to build new URLs using the capabilities of UriComponentsBuilder.
Example:
ServletUriComponentsBuilder builder = ServletUriComponentsBuilder.fromCurrentRequestUri();
builder.scheme("https");
builder.replaceQueryParam("someBoolean", false);
URI newUri = builder.build().toUri();
Java's URI Class can help you out of this:
public static String getCurrentUrl(HttpServletRequest request){
URL url = new URL(request.getRequestURL().toString());
String host = url.getHost();
String userInfo = url.getUserInfo();
String scheme = url.getProtocol();
String port = url.getPort();
String path = request.getAttribute("javax.servlet.forward.request_uri");
String query = request.getAttribute("javax.servlet.forward.query_string");
URI uri = new URI(scheme,userInfo,host,port,path,query,null)
return uri.toString();
}
in jsp file:
request.getAttribute("javax.servlet.forward.request_uri")
You can also add a UriComponentsBuilder to the method signature of your controller method. Spring will inject an instance of the builder created from the current request.
#GetMapping
public ResponseEntity<MyResponse> doSomething(UriComponentsBuilder uriComponentsBuilder) {
URI someNewUriBasedOnCurrentRequest = uriComponentsBuilder
.replacePath(null)
.replaceQuery(null)
.pathSegment("some", "new", "path")
.build().toUri();
//...
}
Using the builder you can directly start creating URIs based on the current request e.g. modify path segments.
See also UriComponentsBuilderMethodArgumentResolver
If you need the URL till hostname and not the path use Apache's Common Lib StringUtil, and from URL extract the substring till third indexOf /.
public static String getURL(HttpServletRequest request){
String fullURL = request.getRequestURL().toString();
return fullURL.substring(0,StringUtils.ordinalIndexOf(fullURL, "/", 3));
}
Example: If fullURL is https://example.com/path/after/url/ then
Output will be https://example.com
System.out.println(((HttpServletRequest)request).getRequestURI());
I used it. hope it's useful.