My problem
In my android application I get url input from user, like "www.google.com".
I want to find out for the given url whether to use http or https.
What I have tried
after referring to some Stack Overflow questions I tried with getScheme()
try {
String url_name="www.google.com";
URI MyUri = new URI(url_name);
String http_or_https="";
http_or_https=MyUri.getScheme();
url_name=http_or_https+"://"+urlname;
Log.d("URLNAME",url_name);
}
catch (Exception e) {
e.printStackTrace();
}
But my above code throws an exception.
My question
Is above approach getScheme() correct or not?
If above approach is incorrect, how to find url http or https?
A domain name is a domain name, it has nothing to do with protocol. The domain is the WHERE, the protocol is the HOW.
The domain is the location you want to go, the protocol is how do you go there, by bus, by plane, by train or by boat. It makes no sense to ask 'I want to go to the store, how do I ask the store if I should go by train or by car?'
The reason this works in browsers is that the browser usually tries to connect using both http and https if no protocol is supplied.
If you know the URL, do this:
public void decideProtocol(URL url) throws IOException {
if ("https".equals(url.getProtocol())) {
// It is https
} else if ("http".equals(url.getProtocol())) {
// It is http
}
}
You can check, the given url is http or https by using URLUtils.
URLUtil.isHttpUrl(String url); returns True if the url is an http.
URLUtil.isHttpsUrl(String url); returns True if the url is an https.
You could use the Apache UrlValidator.
You can specify the allowed url schema and in your case the code look something like this:
String[] schema = {"https"};
UrlValidator urlValidator = new UrlValidator(schemes);
if (urlValidator.isValid("http://foo.bar.com/")) {
System.out.println("url is valid");
} else {
System.out.println("url is invalid");
}
if (urlValidator.isValid("https://foo.bar.com/")) {
System.out.println("url is valid");
} else {
System.out.println("url is invalid");
}
A hierarchical URI is subject to further parsing according to the syntax
[scheme:][//authority][path][?query][#fragment]
so your url_name lack of scheme.
if url_name is "https://www.google.com", so the scheme is https.
refer： http://docs.oracle.com/javase/7/docs/api/java/net/URI.html
You can't get it from just URL. It doesn't make sense. Some websites can work with both http & https. It depends upon website itself whether they use SSL certificate or not.
As was pointed out, if a user didn't dare to provide full url including protocol type. The app might use kind of trial and error approach, try to establish connection using the list of available protocols. (http, https). The successful hit might be considered as a default. Again, all this about usability,this method is better than just annoying an inattentive user with ugly error message.
ranjith- I will store some sort of mapping between URI (main domain) and preferred protcol within application..can have pre-defined mappings based on what you know and then let that mapping grow as more Uris added..
Related
I am trying to find a way to get the protocol from a URL that the user types in. I have an EditText set as uri in an android layout file. The user types in his web address as www.thiersite.com or theirsite.com.
Now how can I get the correct protocol from what they have typed in? It seems everywhere I look that you need to have either https:// or http:// as the protocol in the http request. I get a malformed exception when I don't have a protocol for their web address.
Is there a way to check the URL without having the need to have the protocol when they typed their address? So in essence, do I need to ask the User to type in the protocol as part of the URL? I would prefer to do it programmatically.
/**
* Protocol value could be http:// or https://
*/
boolean usesProtocol(String url,String protocol){
boolean uses = false;
try{
URL u = new URL( protocol.concat(url) );
URLConnection con = u.openConnection();
con.connect();
// the following line will be hit only if the
// supplied protocol is supported
uses = true;
}catch(MalformedURLException e){
// new URL() failed
// user has made a typing error
}catch(IOException e){
// openConnection() failed
// the supplied protocol is not supported
}finally{
return uses;
}
}
I believe that the code is self-explaining. The above code uses no external dependencies. If you do not mind using JSoup, there is another answer on SO that deals with the same: Java how to find out if a URL is http or https?
My Source: http://docs.oracle.com/javase/tutorial/networking/urls/connecting.html
How can I determine if an url is referring to a file or directory. the link http://example.com/test.txt should return that it is a file an http://example.com/dir/ is a directory
I know you can do this with the Uri class but this objects IsFile function only works with the file:/// scheme. And I am working with the http:// scheme. Any idea's?
Thanks
I'm not sure why this question was left unanswered/neglected for a long time. I faced the same situation in server-side java (reckon it would be similar for Android flavour). The only input information is a URL to the resource and we need to tell if the resource is a directory or file. So here's my solution.
if ("file".equals(resourceUrl.getProtocol())
&& new File(resourceUrl.toURI()).isDirectory()) {
// it's a directory
}
Hope this helps the next reader.
Note: please see #awwsmm comment. It was my assumption when provided the answer above. Basically, it doesn't make sense to test if a remote resource is a directory or anything. It is totally up to the site to decide what to return for each request.
It won't work because protocol would be http:// not file://.
class TestURL{
public static boolean isDirectory(URL resourceUrl){
if ("file".equals(resourceUrl.getProtocol())
&& new File(resourceUrl.toURI()).isDirectory()) {
true;
}
return false;
}
public static void main(String[] args){
System.out.println(TestURL.isDirectory("http://example.com/mydir/"));
}
}
I think a directory http://example.com/some/dir will redirect to http://example.com/some/dir/, a file will not.
ie one can examine the http Location field in the HEAD response:
$ curl -I http://example.com/some/dir | grep Location
Location: http://example.com/some/dir/
In my case I needed to download a file with HTTPS and there was occasions that server was misconfigured to redirect the request and thus would download only http data.
In my case I was able to "decide" whether the requested resource was file or not was inspecting "content type" / aka MIME type.
String[] allowedMimeTypes = new String[] { "application/octet-stream", "image/gif",
"text/css", "text/csv", "text/plain", "text/xml" };
URL website = new URL("https://localhost/public/" + file);
HttpsURLConnection huc = (HttpsURLConnection) website
.openConnection();
if (!Arrays.asList(allowedMimeTypes).contains(huc.getContentType())) {
throw new Exception("Not a file...");
}
When the response contained text/html;charset=utf-8 or similar I was able to determine that it was indeed not a file.
Also note that these MIME types are usually quite configurable on server side.
Sample URL http://dineshlingam.appspot.com/guestbook?name=xxx&skills=zzz
Sample Code
public class GuestbookServlet extends HttpServlet
{
public void doGet(HttpServletRequest req, HttpServletResponse res) throws IOException
{
String userName = req.getParameter("name");
String userSkills = req.getParameter("skills");
if(userName != null && userSkills != null)
{
res.setContentType("text/plain");
res.getWriter().println("Name : " + userName + ", Skills : " + userSkills);
}
else
res.sendRedirect(req.getRequestURI());
}
}
I am manually enter this URL to web browser.
How to secure parameters value.
Give me any one suitable example. Because I don't know the java concept and google-app-engine concept.
Really I don't know the SSL. So please I need detailed explanation of SSL with Example.
I am using eclipse to develop my application. Please help me. Thanks.
Your code is a classic example of a page vunerable to a CSS (Cross-Site-Scripting) attack. Using HTTPS wont mitigate that. Instead you need to escape any input before adding it to the page.
For example by using StringEscapeUtils.escapeHtml() and StringEscapeUtils.escapeJavaScript() from the Apache Commons Lang library.
Using https does not secure url parameter by any mean. You have to put parameters either in header or body if you want to make it secure. However if you are making a call directly from browser for this you cant put it in header neither in body because it is a a GET request. +1 to nfechner for highlighting XSS issue in your code.
For your problem here are the possible workaround with https:
Instead of GEt call use a POST call by putting this search in separate form in your page and use HTTPS on top of that.
If you want to use GET request you have to put the parameters in Headers, make a search page, When user hits the search button, make ajax call to above resource by passing it into header using https call.
Hey,
Maybe the title is not the best choice, but I really don't know how to better describe the problem.
The thing is when you point your browser to url that contains #
http://anydomain.com/test/elsem/1234#dogeatdog
and for some reason (ie. there is a business logic) you want to redirect to other page
http://anydomain.com/test/els/1234
the #dogeatdog will be added to new url.
I found this behavior while developing wicket app, but just now I tested it with simple pure java servlet. Can someone explain it to me?
Here is the code just in case I'm doing something wrong:
private void process(HttpServletRequest req, HttpServletResponse res)
{
res.setContentType("text/plain");
try
{
HttpSession session = req.getSession();
Object as = session.getAttribute("as");
if (as == null)
{
log.info("redirecting");
session.setAttribute("as", 1);
res.sendRedirect("/test/");
}
else
{
log.info("writing");
PrintWriter out = res.getWriter();
out.write("after redirect "+as);
out.flush();
}
}
catch (IOException e)
{
e.printStackTrace();
}
}
Hash fragments (#a_hash_fragment) never leave the browser, they are not part of HTTP request.
What the web server gets in this case is GET /test/elsem/1234, and it responds with redirect 3xx code and the new url /test/els/1234, which your browser picks and appends #dogeatdog. Makes sense now?
UPDATE: Thanks to Zack, here's a W3C document that exactly explains how this (should) work:
http://www.w3.org/Protocols/HTTP/Fragment/draft-bos-http-redirect-00.txt
From the sendRedirect Javadoc:
Sends a temporary redirect response to the client using the specified
redirect location URL. This method can accept relative URLs; the
servlet container must convert the relative URL to an absolute URL
before sending the response to the client. 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.
Because of repetitive use of "relative" in the Javadoc, I suspect the new URL is using what it can from the old URL and then building from there...
In the brief amount of what I've read, forwarding should be used if possible instead of redirect.
See this for a good explanation of forward verses redirect.
See this for straight-forward examples of forwarding requests to Servlets or JSPs.
Of course, with forwarding, the original URL will remain intact so that may not be what you're looking for...
EDIT
With information from milan, I found some more information regarding URL fragments (the stuff after "#" - I didn't know that was their official name until corresponding with milan).
There's another SOF post that has some good information concerning this and possibly the best answer: URL Fragment and 302 redirects
I have "+1'd" milan for giving good direction on this...
I am trying to get the domain name from the url with JSTL. The 2 methods I know return the wrong information. I need exactly what is in the URL.
When I do:
${pageContext.request.remoteHost}
I get my server's IP.
When I do:
${pageContext.request.serverName}
I normally get the right domain name, but on a server we have on amazon its returning "server1" instead of the correct domain name, probably because of the way it handles multiple domains.
Anyone know how I can get the current domain name in the URL?
I may need to get the URL and then parse it. How would I do that?
You should be using ServletRequest#getLocalName() instead. It returns the real hostname of the server. The ServletRequest#getServerName() indeed returns the value as set in Host header.
${pageContext.request.localName}
That's all. The solutions suggested in other answers are plain clumsy and hacky.
By the way, the ServletRequest#getRemoteHost() doesn't return the server's hostname, but the client's one (or the IP when the hostname is not immediately resolveable). It's obviously the same as the server's one whenever you run both the webserver and webbrowser at physically the same machine. If you're interested in the server's IP, use ServletRequest#getLocalAddr(). The terms "local" and "remote" must be interpreted from server's perspective, not from the client's. It's after all the server there where all that Java code runs.
You can parse domain name from URL
OR
public static String getDomainName(String url)
{
URL u;
try {
u = new URL(url);
}
catch (Exception e) {
return "";
}
return u.getHost();
}
You can use HttpServletRequest.getRequestUrl() to:
Reconstructs the URL the client used to make the request. The returned URL contains a protocol, server name, port number, and server path, but it does not include query string parameters.
this would return a String like Get domain name in url with JSTL?
It should then be trivial to parse this to find the string that comes after the scheme (http, https, etc) and before the requestURI.
${pageContext.request.contextPath}