I have two websites, A and B. When I open website A, I am redirected to website B automatically.
What is the function with which I can check what was the full path of website A from which was the redirect?
I was trying to start with:
logger.info(request.getPathInfo());
logger.info(request.getPathTranslated());
logger.info(request.getServletPath());
logger.info(request.getLocalName());
logger.info(request.getRemoteAddr());
logger.info(request.getRemoteHost());
logger.info(request.getRequestURI());
logger.info(request.getServerName());
but none of them is correct.
For redirecting I use response.sendRedirect inside Controller.
Thanks for help.
You can try using the optional referer header:
request.getHeader("referer");
But it is important to note that this may not always be populated (specifically IE).
A better solution, if you are in control of both of the websites, is to pass the value somehow when you are doing the redirect. For example, as a GET or POST parameter.
Edit:
As suggested above, you can append query strings to your redirect URL. For example, you might try something like this:
String redirectUrl = "http://my.redirect.com/";
redirectUrl += "?referer=";
redirectUrl += URLEncoder.encode(request.getRequestURL().toString(), "UTF-8");
Then you can just pull this out of the request on the other side.
Use this as a starting point. You may need to manually append other query parameters that may not be part of the getRequestURL() output.
None of these would get you the page that redirected you to the current page. What you can try is:
String refererPage = request.getHeader("referer");
However keep in mind that this is also browser dependent and may not always be present.
Try this
request.getHeader("referer");
Please try
request.getHeader("referer");
Related
I asked this question before and Evgeniy Dorofeev answered it. Although worked for direct link only, but I accepted his answer. He just told me about check the content type from direct link:
String requestUrl = "https://dl-ssl.google.com/android/repository/android-14_r04.zip";
URL url = new URL(requestUrl);
URLConnection c = url.openConnection();
String contentType = c.getContentType();
As far I know, there are two URL types to download a file:
Direct link. For example: https://dl-ssl.google.com/android/repository/android-14_r04.zip. From this link, we can download data directly and get the file name, included with file extension (in this link, .zip extension). So we can know what file to be downloaded. You can try to download from that link.
Undirect link. For example: http://www.example.com/directory/download?file=52378. Have you ever tried to download data from Google Drive? When downloading data from Google Drive, it will gives you an undirect link, such as the link above. We never know whether the link contains a file or webpage. Also, we don't know the file name and file extension is, because of this link type is unclear and random.
I need to check whether it is a file or webpage. I must download it if the content type is a file.
So my question:
How do I check the content type from an undirect link?
As shown in the comments of this question, can HTTP-redirects solves the problem?
Thanks for your help.
After you open an URLConnection, a header file is returned. There are some information about the file in it. You can pull what you want from there. For example:
URLConnection u = url.openConnection();
long length = Long.parseLong(u.getHeaderField("Content-Length"));
String type = u.getHeaderField("Content-Type");
length is size of the file in bytes, type is something like application/x-dosexec or application/x-rar.
Such links redirect browsers to the actual content using HTTP redirects. To get the correct content type, all you have to do is tell HttpURLConnection to follow the redirects by setting setFollowRedirects() to true (documented here).
MimeTypeMap.getFileExtensionFromUrl(url)
This one worked for me, you have to use retrofit to check the headers of response. First you have to define an endpoint to call it with the url you want to check:
#GET
suspend fun getContentType(#Url url: String): Response<Unit>
Then you call it like this to get the content type header:
api.getContentType(url).headers()["content-type"]
If I have a HTML String object, using Selenium in Java, how can I get the browser to open that String as a HTML page? I have seen this done before but I don't remember the format that the URL needs to be.
For this example, let's say the string is :
<h2>This is a <i>test</i></h2>
I looked through this page and couldn't find the answer but I might be overlooking it. For example I tried this URL and it didn't work for me:
data:<h2>This is a <i>test</i></h2>
Here is a link for documentation http://en.wikipedia.org/wiki/Data_URI_scheme. You need to specify MIME-type of data. Try data:text/html,<h2>This is a <i>test</i></h2>
192.168.178.83:18300/Zupdate_we_view.act?name=bunnavitou&position=Research111&sex=Male
hide value on url
from this one :
192.168.178.83:18300/Zupdate_we_view.act?name=bunnavitou&position=Research111&sex=Male
==>To this :
192.168.178.83:18300/Zupdate_we_view.act?
Need Help !! JAVA or JAVA SCRIPT
The best way to hide values on url is to use POST method.
Click HTTP Methods: GET vs. POST or GET vs POST
for more info about GET and POST methods.
First you have to get the url, modify it, and then with javascript replace the "true" url with the one you just modified. Here is what I would do:
//get the url
$old_url = "http://".$_SERVER['HTTP_HOST'].$_SERVER['REQUEST_URI'];
//then you modify it
$new_url = ...;
//then with javascript you replace the url with the new one
<script type="text/javascript">
if(new_url != '')
{
window.history.replaceState({"html":'Title',"pageTitle":'Page Title'}, '', new_url);
}
</script>
This should do the trick, if you really want to use GET ; POST would be better, and you wouldn't have to hide the GET parameters.
Like others suggested, use POST method, but if you want to stick with GET and JavaScript solution is enough (values can be seen in request same as with POST, but furthermore users with disabled JS will see the values), you can use approach described here: https://developer.mozilla.org/en-US/docs/Web/Guide/API/DOM/Manipulating_the_browser_history
Another way is to encode the values on first page & pass those values and decode the value
on second page.
Using that technique you have no need to post data. User can also read the parameter but
it is encoded not the real value.
You can use different algorithm to encode/decode parameter on both side.
If I have a general url (not restricted to twitter or google) like this:
http://t.co/y4o14bI
is there an easy way to check if this url is shortened?
In the above case, I as a human can of course see that it was shortend, but is there an automatic and elegant way?
You could do a request to the URL, look if you get redirected and if so, assume it's a shortening service. For this you'd have to read the HTTP status codes.
On the other hand, you could whitelist some URL shortening services (t.co, bit.ly, and so on) and assume all links to those domains are shortened.
Drawback of the first method is that it isn't certain, some sites use redirects internally. The drawback of the second method is that you'd have to keep adding shortening services, although only a few are used widely.
One signal may be to request the URL and see if it results in a redirect to another domain. However, without a good definition of what "shortened" means, there is no generic way.
if you know all the domains that can be used to shorten your URLs, check if it is contained :
String[] domains = {"bit.ly", "t.co"...};
for(String domain : domains){
if(url.startsWith("http://" + domain)){
return true;
}
}
return false;
You can't: You will have to work by assumption.
Assumption:
Does www exist in url.
Does the server name end with a valid domain (e.g. com, edu, etc.) or does it has co.xx where xx is a valid country or organization code.
And you can add more assumption based on other url shortening links.
You can't.
You can only check if you list a couple of shorteners and check if the url starts with it.
You can also try checking whether the url is shorter than a given length (and contains path/query string), but some shorteners (tinyurl for example) may have longer urls than normal sites (aol.com)
I would prefer the list of known shorteners.
Here's what you could do in Java, groovy and the like.
Get the url you want to test;
Open the url with HttpURLConnection
Check the response code
if it is a valid code, 200 for example, the you can retrieve the url string in long form from the connection object if it was shortened or back in its original form if it wasn't.
We all love to see some code don't we. Its crude, but hey!
String addr = "http://t.co/y4o14bI";
URL url = new URL(addr);
HttpURLConnection connection = (HttpURLConnection) url.openConnection();
if (connection.getResponseCode() == 200) {
String longUrl = connection.url;
System.out.println(longUrl);
} else {
// You decide what you want to do here!
}
Actually, you as a human, can't. The only way you know that it's shortened is that it's a t.co domain. The y4o14bI could be an CMS identifier for all you know.
The best way would be to use a list of known shortener urls, and lookup against that.
And even then you would have problems. I use bit.ly with a personal domain, wtn.gd
So http://wtn.gd/random would also be a shortened URL.
You could maybe do a HTTP HEAD-request, and check for a 301/302 ?
If you request an URL like this, your HttpCLient should receive a HTTP Redirect instead of a HTML page. This wouldn't be an evidence but at least a hint.
Evaluate the URL and look for some clues:
the Path meets certain criteria
only has one step (i.e. not multiple slashes)
does not end with filename extensions
not longer than X characters (would need to evaluate various URL shortening services and adjust the upper bounds for the max token length)
HttpUrlConnection returns a redirect responseCode (i.e. 301, 302)
I would suggest using android.util.Patterns.WEB_URL
public static List<String> findUrls(String input) {
List<String> links = new ArrayList<>();
Matcher m = android.util.Patterns.WEB_URL.matcher(input);
while (m.find()) {
String url = m.group();
links.add(url);
}
return links;
}
Use the unshorten URL service like https://unshorten.me
They have an API as well https://unshorten.me/api
If the URL is shortened it will return the original URL.
If not you will get the same one back.
I have a scenario where the user clicks on a 'restaurant' link (for searching restaurants in a particular locality). I have to check whether the location is set or not. If it is not set, I want to redirect him to a page that allows him to set the location, and, then, go back to search results filtered by the set location. I'm using response.sendRedirect(url) to redirect the user to the setting location page. But, how can I send the redirect back URL (i.e., the URL where I want to send the user after the location is set)?
I tried this:
response.sendRedirect("/location/set.html?action=asklocation&redirectUrl="+
request.getRequestUri()+request.getQueryString());
but this isn't working and 404 error is shown; also, the url formed in the browser doesn't look good.
Please, if anyone could solve the problem ...
Looks like you're missing at least a "?" between request.getRequestUri() and request.getQueryString(). You should url-encode the parameter as well, which you can use java.net.URLEncoder for.
Also, when doing redirects you need to prepend the context path: request.getContextPath().
Something like
String secondRedirectUrl = request.getRequestUri()+"?"+request.getQueryString();
String encodedSecondRedirectUrl = URLEncoder.encode(secondRedirectUrl, serverUrlEncodingPreferablyUTF8);
String firstRedirectUrl = request.getContextPath()+"/location/set.html?action=asklocation&redirectUrl="+encodedSecondRedirectUrl;
response.sendRedirect(firstRedirectUrl);
Personally, i'd rather solve the problem by storing a RequestDispatcher in the session and forwarding to it after the location has been set.
My first response will be to remove the / on your URL, something of this effect (to your code):
response.sendRedirect("location/set.html?action=asklocation&redirectUrl="+
request.getRequestUri()+request.getQueryString());
If that doesn't work, add request.getContextPath() at the beginning of your url string, like so:
response.sendRedirect(request.getContextPath() + "/location/set.html?action=asklocation&redirectUrl="+request.getRequestUri()+request.getQueryString());
The Javadoc states:
If the location is relative without a
leading '/' the container interprets
it as relative to the current request
URI. If the location is relative with
a leading '/' the container interprets
it as relative to the servlet
container root.