How to deal with the URISyntaxException

Question

I got this error message :

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

Answer 1

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");
    }
  }
}

Answer 2

You have to encode your parameters.

Something like this will do:

import java.net.*;
import java.io.*;

public class EncodeParameter { 

    public static void main( String [] args ) throws URISyntaxException ,
                                         UnsupportedEncodingException   { 

        String myQuery = "^IXIC";

        URI uri = new URI( String.format( 
                           "http://finance.yahoo.com/q/h?s=%s", 
                           URLEncoder.encode( myQuery , "UTF8" ) ) );

        System.out.println( uri );

    }
}

http://java.sun.com/javase/6/docs/api/java/net/URLEncoder.html

Answer 3

Use % encoding for the ^ character, viz. http://finance.yahoo.com/q/h?s=%5EIXIC

Answer 4

Replace spaces in URL with + like If url contains dimension1=Incontinence Liners then replace it with dimension1=Incontinence+Liners.