String url = "mysite.com/index.php?name=john&id=432"
How can I extract the id parameter (432)?
it can be in any position in the url and the length of the id varies too
4 Answers
You can use Apache URLEncodedUtils from HttpClient package:
import org.apache.http.NameValuePair;
import org.apache.http.client.utils.URLEncodedUtils;
import java.nio.charset.Charset;
import java.util.List;
public class UrlParsing {
public static void main(String[] a){
String url="http://mysite.com/index.php?name=john&id=42";
List<NameValuePair> args= URLEncodedUtils.parse(url, Charset.defaultCharset());
for (NameValuePair arg:args)
if (arg.getName().equals("id"))
System.out.println(arg.getValue());
}
}
This print 42 to the console.
If you have the url stored in a URI object, you may find useful an overload of URLEncodedUtils.parse that accept directly an URI instance. If you use this overloaded version, you have to give the charset as a string:
URI uri = URI.create("http://mysite.com/index.php?name=john&id=42");
List<NameValuePair> args= URLEncodedUtils.parse(uri, "UTF-8");
3 Comments
124697
it complains about Charset.defaultCharset() it needs a string instead. whats the encoding string look like?
Andrea Parodi
It's working for me. I'm using version 4.2 of httpclient. Beware that my first version was wrong: I edited my code: are you using the code before my edit?
Andrea Parodi
Looked at URLEncodedUtils javadoc: it wnat a String as a second parameter if you give the url as an URI instance; otherwise, it accept a Charset as in my code
I just give an abstract regex. add anything you don't want in id after [^&
Pattern pattern = Pattern.compile("id=([^&]*?)$|id=([^&]*?)&");
Matcher matcher = pattern.matcher(url);
if (matcher.matches()) {
int idg1 = Integer.parseInt(matcher.group(1));
int idg2 = Integer.parseInt(matcher.group(2));
}
either idg1 or idg2 has value.
Comments
You can use:
String id = url.replaceAll("^.*?(?:\\?|&)id=(\\d+)(?:&|$).*$", "$1");
5 Comments
Affe
fails if it's the first param :)
124697
that returns the id and the rest of the url (the other parameters). output 432&pId=10....
anubhava
@user521180: So sorry, omitted the
.*$ part earlier, pls check the edited answer now.anubhava
@user521180: Are you sure, what is your input URL? For me it always returns parameter
id's value.
Kevin Day
wow - and people have trouble reading regex and figuring out what they do :-)
The regex has already been given, but you can do it with some simple splitting too:
public static String getId(String url) {
String[] params = url.split("\\?");
if(params.length==2) {
String[] keyValuePairs = params[1].split("&");
for(String kvp : keyValuePairs) {
String[] kv = kvp.split("=");
if(kv[0].equals("id")) {
return kv[1];
}
}
}
throw new IllegalStateException("id not found");
}
id