Develop Reference

Get domain from strange yet valid url with java

I need to get host from this url
android-app://com.google.android.googlequicksearchbox?Pub_id={siteID}
java.net.URL and java.net.URI can't handle it.

The problem is in { and } characters which are not valid for URI. Looks like a placeholder that wasn't resolved correctly when creating a URI.
You can use String.replaceAll() to get rid of these two characters:
String value = "android-app://com.google.android.googlequicksearchbox?Pub_id={siteID}";
URI uri = URI.create(value.replaceAll("[{}]", ""));
System.out.println(uri.getHost()); // com.google.android.googlequicksearchbox

You see, eventually I need path, scheme and query.
I've just found super fast library for parsing such URLs. https://github.com/anthonynsimon/jurl
It's also very flexible.

You can try the following code
String url = "android-app://com.google.android.googlequicksearchbox?Pub_id={siteID}";
url = url.replace("{", "").replace("}","");
URI u;
try {
u = new URI(url);
System.out.println(u.getHost());
} catch (URISyntaxException e) {
e.printStackTrace();
}

java or node.js removes one character when sending or receiving a string

I'm trying to send a string from a java (android) app to a node.js server.
But one character disappears somewhere in the middle and I can't really figure out why.
To send I use a HttpUrlConnection (conn) and send the string like this:
try {
OutputStream os = conn.getOutputStream();
os.write(json.getBytes());
os.close();
} catch (Exception e) {
e.printStackTrace();
}
Here is the base64 encoded string when sent, and string when received:
khVGUBH2kNAR5PPRy7v5dO5iz48Rc7benYARu78\/9wY=\n
khVGUBH2kNAR5PPRy7v5dO5iz48Rc7benYARu78/9wY=\n
so one backslash has be removed.
In node I use this:
exports.getString = function(req, res) {
var string = req.body.thestring;
}
which outputs the later of the two strings.
var express = require('express'),
http = require('http'),
stylus = require('stylus'),
nib = require('nib');
var app = express();
app.configure(function () {
app.use(express.logger('dev'));
//app.use(express.bodyParser());
app.use(express.json());
app.use(express.urlencoded());
app.use(app.router);
}
Any ideas of how I can get the missing character?

The missing backslash character is most probably disappearing in node.js side.
As per the chosen answer on the following question:
Two part question on JavaScript forward slash
As far as JS is concerned / and \ / are identical inside a string
So maybe a fix from Java's would solve your problem by using String's replaceAll method to replace all occurrences of \/ with \\/:
os.write(json.replaceAll("\\/", "\\\\/").getBytes());
Note that replaceAll returns the new string and doesn't change the original string.

Making the base64 encoding url safe solved my problem.

how to avoid recursive url-decoding

I am writing something like an image proxy where I receive URLs from my site front-end, and I download images , re-size them, and return smaller images for the front end and client to download from the "proxy".
This means I need to take care of all-sorts of url patterns, this is why I chose to decode the given url and than encode it using URIUtils.decode:
private String fixUrl(String fromUrl) throws URIException {
fromUrl = URIUtil.decode(fromUrl);
fromUrl = URIUtil.encodeQuery(fromUrl);
return fromUrl;
}
This should help me take care of urls that are already encoded.
My problem is that some of the urls are double encoded, and from what I saw, URIUtils.decode, performs recursive decode and this means that in cases of double encoded urls I will get a bad url that does not work.
Is there a simple way to decode only once?

I'd try to check that URL still contains character %. If it does not contain any % it is not encoded and you can stop your decoding procedure.

Easiest option I know is to use the built-in
java.net.URLDecoder.decode
In order to decode automatically only once.
If like me you have the case that sometimes URL's are double / triple encoded - you can use this recursive function in order to decode again and again until there are no "%" or "+" :
private static String completeDecode(String url) {
if(url.contains("%") || url.contains("+"))
{
try
{
return(completeDecode(java.net.URLDecoder.decode(url, "UTF-8")));
}
catch (UnsupportedEncodingException e)
{
e.printStackTrace();
}
}
return url;
}
Cheers

Using boilerpipe to extract non-english articles

I am trying to use boilerpipe java library, to extract news articles from a set of websites.
It works great for texts in english, but for text with special characters, for example, words with accent marks (história), this special characters are not extracted correctly. I think it is an encoding problem.
In the boilerpipe faq, it says "If you extract non-English text you might need to change some parameters" and then refers to a paper. I found no solution in this paper.
My question is, are there any params when using boilerpipe where i can specify the encoding? Is there any way to go around and get the text correctly?
How i'm using the library:
(first attempt based on the URL):
URL url = new URL(link);
String article = ArticleExtractor.INSTANCE.getText(url);
(second on the HTLM source code)
String article = ArticleExtractor.INSTANCE.getText(html_page_as_string);

You don't have to modify inner Boilerpipe classes.
Just pass InputSource object to the ArticleExtractor.INSTANCE.getText() method and force encoding on that object. For example:
URL url = new URL("http://some-page-with-utf8-encodeing.tld");
InputSource is = new InputSource();
is.setEncoding("UTF-8");
is.setByteStream(url.openStream());
String text = ArticleExtractor.INSTANCE.getText(is);
Regards!

Well, from what I see, when you use it like that, the library will auto-chose what encoding to use. From the HTMLFetcher source:
public static HTMLDocument fetch(final URL url) throws IOException {
final URLConnection conn = url.openConnection();
final String ct = conn.getContentType();
Charset cs = Charset.forName("Cp1252");
if (ct != null) {
Matcher m = PAT_CHARSET.matcher(ct);
if(m.find()) {
final String charset = m.group(1);
try {
cs = Charset.forName(charset);
} catch (UnsupportedCharsetException e) {
// keep default
}
}
}
Try debugging their code a bit, starting with ArticleExtractor.getText(URL), and see if you can override the encoding

Ok, got a solution.
As Andrei said, i had to change the class HTMLFecther, which is in the package de.l3s.boilerpipe.sax
What i did was to convert all the text that was fetched, to UTF-8.
At the end of the fetch function, i had to add two lines, and change the last one:
final byte[] data = bos.toByteArray(); //stays the same
byte[] utf8 = new String(data, cs.displayName()).getBytes("UTF-8"); //new one (convertion)
cs = Charset.forName("UTF-8"); //set the charset to UFT-8
return new HTMLDocument(utf8, cs); // edited line

Boilerpipe's ArticleExtractor uses some algorithms that have been specifically tailored to English - measuring number of words in average phrases, etc. In any language that is more or less verbose than English (ie: every other language) these algorithms will be less accurate.
Additionally, the library uses some English phrases to try and find the end of the article (comments, post a comment, have your say, etc) which will clearly not work in other languages.
This is not to say that the library will outright fail - just be aware that some modification is likely needed for good results in non-English languages.

Java:
import java.net.URL;
import org.xml.sax.InputSource;
import de.l3s.boilerpipe.extractors.ArticleExtractor;
public class Boilerpipe {
public static void main(String[] args) {
try{
URL url = new URL("http://www.azeri.ru/az/traditions/kuraj_pehlevanov/");
InputSource is = new InputSource();
is.setEncoding("UTF-8");
is.setByteStream(url.openStream());
String text = ArticleExtractor.INSTANCE.getText(is);
System.out.println(text);
}catch(Exception e){
e.printStackTrace();
}
}
}
Eclipse:
Run > Run Configurations > Common Tab. Set Encoding to Other(UTF-8), then click Run.

I had the some problem; the cnr solution works great. Just change UTF-8 encoding to ISO-8859-1. Thank's
URL url = new URL("http://some-page-with-utf8-encodeing.tld");
InputSource is = new InputSource();
is.setEncoding("ISO-8859-1");
is.setByteStream(url.openStream());
String text = ArticleExtractor.INSTANCE.getText(is);

Java - Convert String to valid URI object

I am trying to get a java.net.URI object from a String. The string has some characters which will need to be replaced by their percentage escape sequences. But when I use URLEncoder to encode the String with UTF-8 encoding, even the / are replaced with their escape sequences.
How can I get a valid encoded URL from a String object?
http://www.google.com?q=a b gives http%3A%2F%2www.google.com... whereas I want the output to be http://www.google.com?q=a%20b
Can someone please tell me how to achieve this.
I am trying to do this in an Android app. So I have access to a limited number of libraries.

You might try: org.apache.commons.httpclient.util.URIUtil.encodeQuery in Apache commons-httpclient project
Like this (see URIUtil):
URIUtil.encodeQuery("http://www.google.com?q=a b")
will become:
http://www.google.com?q=a%20b
You can of course do it yourself, but URI parsing can get pretty messy...

Android has always had the Uri class as part of the SDK:
http://developer.android.com/reference/android/net/Uri.html
You can simply do something like:
String requestURL = String.format("http://www.example.com/?a=%s&b=%s", Uri.encode("foo bar"), Uri.encode("100% fubar'd"));

I'm going to add one suggestion here aimed at Android users. You can do this which avoids having to get any external libraries. Also, all the search/replace characters solutions suggested in some of the answers above are perilous and should be avoided.
Give this a try:
String urlStr = "http://abc.dev.domain.com/0007AC/ads/800x480 15sec h.264.mp4";
URL url = new URL(urlStr);
URI uri = new URI(url.getProtocol(), url.getUserInfo(), url.getHost(), url.getPort(), url.getPath(), url.getQuery(), url.getRef());
url = uri.toURL();
You can see that in this particular URL, I need to have those spaces encoded so that I can use it for a request.
This takes advantage of a couple features available to you in Android classes. First, the URL class can break a url into its proper components so there is no need for you to do any string search/replace work. Secondly, this approach takes advantage of the URI class feature of properly escaping components when you construct a URI via components rather than from a single string.
The beauty of this approach is that you can take any valid url string and have it work without needing any special knowledge of it yourself.

Even if this is an old post with an already accepted answer, I post my alternative answer because it works well for the present issue and it seems nobody mentioned this method.
With the java.net.URI library:
URI uri = URI.create(URLString);
And if you want a URL-formatted string corresponding to it:
String validURLString = uri.toASCIIString();
Unlike many other methods (e.g. java.net.URLEncoder) this one replaces only unsafe ASCII characters (like ç, é...).
In the above example, if URLString is the following String:
"http://www.domain.com/façon+word"
the resulting validURLString will be:
"http://www.domain.com/fa%C3%A7on+word"
which is a well-formatted URL.

If you don't like libraries, how about this?
Note that you should not use this function on the whole URL, instead you should use this on the components...e.g. just the "a b" component, as you build up the URL - otherwise the computer won't know what characters are supposed to have a special meaning and which ones are supposed to have a literal meaning.
/** Converts a string into something you can safely insert into a URL. */
public static String encodeURIcomponent(String s)
{
StringBuilder o = new StringBuilder();
for (char ch : s.toCharArray()) {
if (isUnsafe(ch)) {
o.append('%');
o.append(toHex(ch / 16));
o.append(toHex(ch % 16));
}
else o.append(ch);
}
return o.toString();
}
private static char toHex(int ch)
{
return (char)(ch < 10 ? '0' + ch : 'A' + ch - 10);
}
private static boolean isUnsafe(char ch)
{
if (ch > 128 || ch < 0)
return true;
return " %$&+,/:;=?#<>#%".indexOf(ch) >= 0;
}

You can use the multi-argument constructors of the URI class. From the URI javadoc:
The multi-argument constructors quote illegal characters as required by the components in which they appear. The percent character ('%') is always quoted by these constructors. Any other characters are preserved.
So if you use
URI uri = new URI("http", "www.google.com?q=a b");
Then you get http:www.google.com?q=a%20b which isn't quite right, but it's a little closer.
If you know that your string will not have URL fragments (e.g. http://example.com/page#anchor), then you can use the following code to get what you want:
String s = "http://www.google.com?q=a b";
String[] parts = s.split(":",2);
URI uri = new URI(parts[0], parts[1], null);
To be safe, you should scan the string for # characters, but this should get you started.

I had similar problems for one of my projects to create a URI object from a string. I couldn't find any clean solution either. Here's what I came up with :
public static URI encodeURL(String url) throws MalformedURLException, URISyntaxException
{
URI uriFormatted = null;
URL urlLink = new URL(url);
uriFormatted = new URI("http", urlLink.getHost(), urlLink.getPath(), urlLink.getQuery(), urlLink.getRef());
return uriFormatted;
}
You can use the following URI constructor instead to specify a port if needed:
URI uri = new URI(scheme, userInfo, host, port, path, query, fragment);

Well I tried using
String converted = URLDecoder.decode("toconvert","UTF-8");
I hope this is what you were actually looking for?

The java.net blog had a class the other day that might have done what you want (but it is down right now so I cannot check).
This code here could probably be modified to do what you want:
http://svn.apache.org/repos/asf/incubator/shindig/trunk/java/common/src/main/java/org/apache/shindig/common/uri/UriBuilder.java
Here is the one I was thinking of from java.net: https://urlencodedquerystring.dev.java.net/

Or perhaps you could use this class:
http://developer.android.com/reference/java/net/URLEncoder.html
Which is present in Android since API level 1.
Annoyingly however, it treats spaces specially (replacing them with + instead of %20). To get round this we simply use this fragment:
URLEncoder.encode(value, "UTF-8").replace("+", "%20");

I ended up using the httpclient-4.3.6:
import org.apache.http.client.utils.URIBuilder;
public static void main (String [] args) {
URIBuilder uri = new URIBuilder();
uri.setScheme("http")
.setHost("www.example.com")
.setPath("/somepage.php")
.setParameter("username", "Hello Günter")
.setParameter("p1", "parameter 1");
System.out.println(uri.toString());
}
Output will be:
http://www.example.com/somepage.php?username=Hello+G%C3%BCnter&p1=paramter+1