How to deal with the URISyntaxException

Frank picture Frank · Apr 15, 2009 · Viewed 115.8k times · Source

I got this error message :

java.net.URISyntaxException: Illegal character in query at index 31: http://finance.yahoo.com/q/h?s=^IXIC

My_Url = http://finance.yahoo.com/q/h?s=^IXIC

When I copied it into a browser address field, it showed the correct page, it's a valid URL, but I can't parse it with this: new URI(My_Url)

I tried : My_Url=My_Url.replace("^","\\^"), but

  1. It won't be the url I need
  2. It doesn't work either

How to handle this ?

Frank

Answer

Eddie picture Eddie · Apr 15, 2009

You need to encode the URI to replace illegal characters with legal encoded characters. If you first make a URL (so you don't have to do the parsing yourself) and then make a URI using the five-argument constructor, then the constructor will do the encoding for you.

import java.net.*;

public class Test {
  public static void main(String[] args) {
    String myURL = "http://finance.yahoo.com/q/h?s=^IXIC";
    try {
      URL url = new URL(myURL);
      String nullFragment = null;
      URI uri = new URI(url.getProtocol(), url.getHost(), url.getPath(), url.getQuery(), nullFragment);
      System.out.println("URI " + uri.toString() + " is OK");
    } catch (MalformedURLException e) {
      System.out.println("URL " + myURL + " is a malformed URL");
    } catch (URISyntaxException e) {
      System.out.println("URI " + myURL + " is a malformed URL");
    }
  }
}