Solved

get a value from a content string

Posted on 2004-04-25
10
287 Views
Last Modified: 2010-03-31
xmlurl=http%3A%2F%2F127.0.0.1%2Fmydbs%2Fwipdesign.nsf%2FagtGetDocsViewXML%3Fopenagent%26u%3Dc956395%26uname%3DJoe+Blogg%26start%3D1%26count%3D9999%26viewname%3Dfrmgicsissueview%26frmname%3D%26frmtype%3DView&xslurl=http%3A%2F%2F127.0.0.1%2Fmydbs%2Fwipdesign.nsf%2Fviewreport.xsl

I do have a method to decode the content string, however I need a method to pass in this content string and get the values for xmlurl and xslurl.

Thanks
pcorreya
0
Comment
Question by:pcorreya
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 3
  • 2
  • 2
  • +1
10 Comments
 
LVL 7

Accepted Solution

by:
vbandaru earned 43 total points
ID: 10911717
The String Object has an overloaded method called indexOf. You should be able to use that in conjunction with substring to get the xmlurl and xslurl.

something like this.

String contentString="xmlurl=http%3A%2F%2F127.0.0.1%2Fmydbs%2Fwipdesign.nsf%2FagtGetDocsViewXML%3Fopenagent%26u%3Dc956395%26uname%3DJoe+Blogg%26start%3D1%26count%3D9999%26viewname%3Dfrmgicsissueview%26frmname%3D%26frmtype%3DView&xslurl=http%3A%2F%2F127.0.0.1%2Fmydbs%2Fwipdesign.nsf%2Fviewreport.xsl";

String xmlurl= contentString.substring(7,contentString.indexOf("xslurl")-1);

String xslurl= contentString.substring(contentString.indexOf("xslurl")+7);
0
 
LVL 86

Assisted Solution

by:CEHJ
CEHJ earned 41 total points
ID: 10912076
Or you can use regular expressions:

            String x = "xmlurl=http%3A%2F%2F127.0.0.1%2Fmydbs%2Fwipdesign.nsf%2FagtGetDocsViewXML%3Fopenagent%26u%3Dc956395%26uname%3DJoe+Blogg%26start%3D1%26count%3D9999%26viewname%3Dfrmgicsissueview%26frmname%3D%26frmtype%3DView&xslurl=http%3A%2F%2F127.0.0.1%2Fmydbs%2Fwipdesign.nsf%2Fviewreport.xsl";
            x = URLDecoder.decode(x, "UTF-8");
            final String RE = "xmlurl=(.+)\\?.+xslurl=(.+)";
            Pattern p = Pattern.compile(RE);
            Matcher m = p.matcher(x);
            if (m.matches()) {
                  System.out.println(m.group(1));
                  System.out.println(m.group(2));
            }
0
 
LVL 92

Expert Comment

by:objects
ID: 10914093
vbandaru is right, the only additional thing you may want to do is decode the urls:

import java.net.*;

public class extracturls
{
      public static void main(String[] args)
      {
            String contentString="xmlurl=http%3A%2F%2F127.0.0.1%2Fmydbs%2Fwipdesign.nsf%2FagtGetDocsViewXML%3Fopenagent%26u%3Dc956395%26uname%3DJoe+Blogg%26start%3D1%26count%3D9999%26viewname%3Dfrmgicsissueview%26frmname%3D%26frmtype%3DView&xslurl=http%3A%2F%2F127.0.0.1%2Fmydbs%2Fwipdesign.nsf%2Fviewreport.xsl";

            String xmlurl= URLDecoder.decode(contentString.substring(7,contentString.indexOf("xslurl")-1));

            String xslurl= URLDecoder.decode(contentString.substring(contentString.indexOf("xslurl")+7));
            
            System.out.println(xmlurl);
            System.out.println(xslurl);
      }
}
0
Technology Partners: We Want Your Opinion!

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

 
LVL 86

Expert Comment

by:CEHJ
ID: 10914109
>>vbandaru is right

? So i'm wrong am i? ;-)

In any case, your comment is redundant - you obviously have not read the question:

>>I do have a method to decode the content string, ...
0
 
LVL 1

Author Comment

by:pcorreya
ID: 10914501
Thanks you all for your contribution, however it would be nice to have a class that I can use to pass it the content string and the lookup field and return its value.
0
 
LVL 92

Assisted Solution

by:objects
objects earned 41 total points
ID: 10914808
try something like:

     public static String extractField(String contentString, String lookup)
     {
          String[] pairs = contentString.split("&");
          for (int i=0; i<pairs.length; i++)
          {
              String[] keyvalue = pairs[i].split("=");
              if (keyvalue[0].equals(lookup)) return keyvalue[1];
          }
          return null;
     }
0
 
LVL 1

Author Comment

by:pcorreya
ID: 10927241
objects, i am getting the message Method split(java.lang.String) not found in class java.lang.String
0
 
LVL 92

Expert Comment

by:objects
ID: 10927304
it only became available in 1.4.
you'll need to use StringTokenizer in previous versions.
0

Featured Post

The Ultimate Checklist to Optimize Your Website

Websites are getting bigger and complicated by the day. Video, images, custom fonts are all great for showcasing your product/service. But the price to pay in terms of reduced page load times and ultimately, decreased sales, can lead to some difficult decisions about what to cut.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Java had always been an easily readable and understandable language.  Some relatively recent changes in the language seem to be changing this pretty fast, and anyone that had not seen any Java code for the last 5 years will possibly have issues unde…
Go is an acronym of golang, is a programming language developed Google in 2007. Go is a new language that is mostly in the C family, with significant input from Pascal/Modula/Oberon family. Hence Go arisen as low-level language with fast compilation…
Viewers learn how to read error messages and identify possible mistakes that could cause hours of frustration. Coding is as much about debugging your code as it is about writing it. Define Error Message: Line Numbers: Type of Error: Break Down…
Viewers will learn one way to get user input in Java. Introduce the Scanner object: Declare the variable that stores the user input: An example prompting the user for input: Methods you need to invoke in order to properly get  user input:

690 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question